# Three Dimensional Geometry Class 12 Mathematics Important Questions

Please refer to Three Dimensional Geometry Class 12 Mathematics Important Questions with solutions provided below. These questions and answers have been provided for Class 12 Mathematics based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 12 Mathematics for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 12.

## Class 12 Mathematics Important Questions Three Dimensional Geometry

Question. If a line makes angles 90°, 60° and θ with x, y and z-axis respectively, where θ is acute, then find θ.
Answer. Since, cos2α + cos2β + cos2γ = 1
⇒ cos2 90° + cos2 60° + cos2 θ = 1

Question. If a line makes angles α, β, γ with the positive direction of coordinate axes, then write the value of sin2α + sin2β + sin2γ.
Answer. Here, the direction cosines of the given line are cos α, cos β, cos γ and cos2α + cos2β + cos2γ = 1
⇒ (1 – sin2α) + (1 – sin2β) + (1 – sin2γ) = 1
⇒ sin2α + sin2β + sin2γ = 2.

Question. If a line has direction ratios 2, –1, –2, then what are its direction cosines?
Answer. The direction ratios of the given line are 2, –1, –2.
Let a = 2, b = –1, c = –2

Question. Find the sum of the intercepts cut off by the plane 2x + y – z = 5, on the coordinate axes.
Answer. We have, 2x + y – z = 5

Question. Write the vector equation of the line passing though (1, 2, 3) and perpendicular to the plane r̅ (î + 2ĵ − 5) + 9 = 0.
r̅ · (î + 2ĵ − 5k̂) + 9 = 0                                         …(i)
⇒ (xî + yjˆ + zk̂ ) · (î + 2 jˆ − 5k̂ ) + 9 = 0
⇒ x + 2y – 5z + 9 = 0
∴  Equation of line through(1, 2, 3) and ⊥ to the plane (i) is

Question. Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane (î + ĵ + k̂) = 2
Answer. The direction ratios of normal to the given plane r̅ (î + ĵ + k̂) = 2 are 1, 1, 1 .
So, cartesian equation of plane passing through (a, b, c) and having d.r’s of normal proportional to 1, 1, 1 is
⇒ 1·(x – a) + 1· (y – b) + 1· (z – c) = 0
⇒ x + y + z = a + b + c
∴ Vector equation of the required plane is,
r̅ (î + ĵ + k̂) = a + b + c

Question. Write the direction-cosines of the line joining the points (1, 0, 0) and (0, 1, 1).
Answer. The direction-cosines of the line joining the points (1, 0, 0) and (0, 1, 1) are

Question. Write the direction cosines of a line equally inclined to the three coordinate axes.
Answer. Here, line is equally inclined to the axes.
∴ l = m = n. We know that, l2 + m2 + n2 = 1

Question. The equation of a line are 5x – 3 = 15y + 7 = 3 –10z. Write the direction cosines of the line.
Answer. The given line is 5x – 3 = 15y + 7 = 3 –10z

Question. Find the shortest distance between the two lines whose vector equations are

Question. Find the shortest distance between the two lines whose vector equations are

Question. By computing the shortest distance between the following pair of lines, determine whether they intersect or not?

Answer. The eqns. of the lines are

Question. Find the shortest distance between the following pair of lines:

Question. Find the shortest distance between the following lines whose vector equations are

Answer. From the given equations, we get

Question. Find the unit vector perpendicular to the plane ABC where the position victors of A, B and C are 2î − ĵ + k̂, î + ĵ + 2k̂ and 2î + 3k̂ respectively.
Answer. Intercepts are a = 3, b = –4, c = 2
∴ The intercept form of the plane is

Question. Find the equation of the plane passing through the intersection of the planes r̅ · (2î + ĵ + 3k̂) = 7, r̅ · (2î + 5ĵ + 3k̂) = 9 and the point (2, 1, 3).

Question.

Question. Find the vector equation of the plane through the points (2, 1, –1) and (–1, 3, 4) and perpendicular to the plane x – 2y + 4z = 10.
Answer. Let the eq. of the plane through (2, 1, –1) be
a(x – 2) + b(y – 1) + c(z + 1) = 0                               …(i)
Also, point (–1, 3, 4) lies on it
∴ –3a + 2b + 5c = 0                               …(ii)
Also (i) is ⊥ to the plane x – 2y + 4z = 10
∴ a · 1 + b(–2) + c · 4 = 0
⇒ a – 2b + 4c = 0                               …(iii)
Solving (ii) and (iii), we get

∴ From (i), required eqn. of the plane is
18(x – 2) + 17(y – 1) + 4(z + 1) = 0
⇒ 18x + 17y + 4z = 49
In vector form, eq. of this plane is
r̅ ·  (18î +17ĵ + 4k̂) = 49.

Question. Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y –3z = 8.
Answer. The equation of plane passing through (1, –1, 2) is
a(x – 1) + b(y + 1) + c(z – 2) = 0                               …(i)
If plane (i) is perpendicular to each of the planes
2x + 3y – 2z = 5 and x + 2y – 3z = 8, then
2a + 3b – 2c = 0                                     …(ii)
and a + 2b – 3c = 0                               …(iii)
Solving (ii) and (iii), we get

Putting the values of a, b and c in (i), we get
–5λ(x – 1) + 4λ(y + 1) + λ(z – 2) = 0
⇒ –5x + 5 + 4y + 4 + z – 2 = 0
⇒ 5x – 4y – z – 7 = 0

Question. Find the distance between the point (–1, –5, –10) and the point of intersection of

Answer. Equation of given line is

Question. Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, –4, –5) and B(2, –3, 1) intersects the plane 2x + y + z = 7.
Answer. Eq. of the line joining the points A(3, –4, –5) and B(2, – 3, 1) is.

Question. A plane makes intercepts –6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.
Answer. Given that –6, 3, 4 are intercepts on x, y and z-axes respectively.

Question.

Answer. Any point on the line

Question.

⇒ 3λ + 1 = 2μ + 4, 1 – λ = 0 and –1 = 3μ –1
On solving last two equations, we get λ = 1 and μ = 0.
These values of λ and μ satisfy the first equation.
So, the given lines intersect.
Putting λ = 1 in (i), we get the position vector of the point of intersection.
Thus, the coordinates of the point of intersection are (4, 0, –1).

Question. Show that the lines

If these lines intersect, they must have a common point. So, we must have

∴ The given lines intersect and their point of intersection is (–1, –6, –12) .

Question. Find the image of the point having position vector î + 3ĵ + 4k̂ in the plane r̅ · (2î − ĵ + k̂) + 3 = 0 .
Answer. Let P be the point with position vector î + 3ĵ + 4k̂
≡ (1, 3, 4) and the plane is r̅ ·(2î − ĵ + k̂) + 3 = 0
⇒ 2x – y + z + 3 = 0                                                         …(i)
Any line ⊥ to this plane passes through P(1, 3, 4) is

Question. Find the equation of the plane passing through the intersection of the planes x + 3y + 6 = 0 and 3x – y – 4z = 0 and whose perpendicular distance from the origin is unity.
Answer. Equation of plane passing through intersection of given planes is
x + 3y + 6 + λ(3x – y – 4z) = 0

Question. Find the coordinates of the point, where the line

intersects the plane x – y + z – 5 = 0. Also find the angle between the line and the plane.
Answer. The given equation of line is

and equation of plane is
x – y + z – 5 = 0 …(ii)
From (i), we have
x = 3λ + 2, y = 4λ – 1, z = 2λ + 2
Line (i) intersects the plane, so the point
(3λ + 2, 4λ – 1, 2λ + 2) will also lie on plane.
∴ (3λ + 2) – (4λ – 1) + (2λ + 2) – 5 = 0
⇒ λ + 0 = 0 ⇒ λ = 0
So, the required coordinates of the point are (2, –1, 2).
Now the direction ratios of line (i) are 3, 4, 2 and direction ratios of plane (ii) are 1, –1, 1. So, the angle between line (i) and plane (ii) is

Question. Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XY-plane.
Answer. The eq. of the line through A(3, 4, 1) and

Question. Find the coordinates of the point where the line through the points (3, –4, –5) and (2, –3, 1) crosses the plane 2x + y + z = 7.
Answer. Eq. of the line through the points (3, –4, –5) and (2, –3, 1) is

Any point on (i) is (–k + 2, k –3, 6k + 1).
Suppose it lies on the plane 2x + y + z = 7
⇒ 2(–k + 2) + k – 3 + 6k + 1 = 7
⇒ 5k = 5 ⇒ k = 1
∴ The required point is (1, –2, 7)

Question. Find the cartesian equation of the plane passing through the points A (0, 0, 0) and B(3, –1, 2) and parallel to the line

Answer. Equation of plane passing through (0, 0, 0) is
a(x) + b(y) + c(z) = 0                          …(i)
It also passes through (3, –1, 2)
∴ 3a – b + 2c = 0                               …(ii)

Question. Find the coordinates of the point where the line

Substituting the values of x, y, z in the plane
x + y + 4z = 6, we get
2λ – 1 + 3λ – 2 + 4(4λ – 3) = 6
⇒ 21λ – 15 = 6 ⇒ λ = 1
Putting the value of λ in (i), we get
x = 1, y = 1, z = 1
∴ The coordinates of the point where given line
meets the given plane are (1, 1, 1).

Question. A line makes angle α, β, γ and δ with the diagonals of a cube. Prove that,
Answer. A cube is a rectangular parallelopiped having equal length, breadth and height. Let OADBFEGC be the cube with each side of length a units.

The four diagonals are OE, AF, BG and CD.
The direction cosines of the diagonal OE which is the line joining two points O and E are

Question. Find the coordinates of the foot of perpendicular and the length of the perpendicular drawn from the point P(5, 4, 2) to the line

Any point Q on (i) is (2λ – 1, 3λ + 3, – λ + 1)
Also, the given point is P(5, 4, 2).
Now d.r’s of the line PQ are
(2λ – 1 – 5, 3λ + 3 – 4, –λ + 1 – 2)
= (2λ – 6, 3λ – 1, –λ –1).
For PQ to be ⊥ to (i), we must have
(2λ – 6). 2 + (3λ – 1) . 3 + (– λ – 1). (–1) = 0
⇒ 14λ – 14 = 0 ⇒ λ = 1
∴ Q is (1, 6, 0)
which is the foot of ⊥ from P on line (i).

Question. The points A(4, 5, 10), B(2, 3, 4) and C(1, 2, –1) are three vertices of a parallelogram ABCD. Find the vector equations of sides AB and BC and also find the coordinates of point D.
Answer. Let the coordinates of D be (x, y, z).
Vector equation of side AB is

Question. Find the image of the point (1, 6, 3) in the line

Also, write the equation of the line joining the given point and its image and find the length of the segment joining the given point and its image.
Answer. Let Q (x1, y1, z1) be the image of the point P(1, 6, 3) and R be the foot of perpendicular from P to the given line

Let the coordinates of R be (λ, 2λ + 1, 3λ + 2).
So, the direction ratios of PR are
λ – 1, 2λ + 1 – 6, 3λ + 2 – 3
i.e., λ – 1, 2λ – 5, 3λ – 1
∴ PR is perpendicular to the given line.
∴ 1(λ – 1) + 2(2λ – 5) + 3(3λ – 1) = 0
⇒ λ – 1 + 4λ – 10 + 9λ – 3 = 0
⇒ 14λ = 14 ⇒ λ = 1
∴ Coordinates of R are (1, 3, 5)
Since, R is the mid point of P and Q.

Question. Find the equation of a line passing through the points A(0, 6, –9) and B(–3, –6, 3). If D is the foot of perpendicular drawn from a point C(7, 4, –1) on the line AB, then find the coordinates of the point D and the equation of line CD.
Answer. Equation of line passing through A(0, 6, –9)

Any point on this line is (λ, 4λ + 6, – 4λ – 9)
Let coordinates of D be (λ, 4λ + 6, – 4λ – 9)
Coordinates of C are (7, 4, –1).
∴ Direction ratios of CD are
λ – 7, 4λ + 6 – 4, – 4λ – 9 + 1 or λ – 7, 4λ + 2, – 4λ – 8
CD is perpendicular to line AB.
∴ λ – 7 + 4(4λ + 2) – 4(– 4λ – 8) = 0
⇒ λ – 7 + 16λ + 8 + 16λ + 32 = 0
⇒ 33λ + 33 = 0 ⇒ λ = – 1
So, coordinates of point D are (– 1, 2, – 5).
∴ Equation of CD is

Question. Find the perpendicular distance of the point (2, 3, 4) from the line

Also, find the coordinates of the foot of the perpendicular
Answer. Let foot of perpendicular from P(2, 3, 4) on

Any point on the line is (– 2λ + 4, 6λ, – 3λ + 1)
Let coordinates of Q be (– 2λ + 4, 6λ, – 3λ + 1)
Direction ratios of PQ are
– 2λ + 4 – 2, 6λ – 3, – 3λ + 1 – 4
or – 2λ + 2, 6λ – 3, – 3λ – 3
PQ is perpendicular to given line so
– 2(– 2λ + 2) + 6(6λ – 3) – 3(– 3λ – 3) = 0
⇒ 4λ – 4 + 36λ – 18 + 9λ + 9 = 0
⇒ 49 −13 = 0 ⇒ λ = 13/49

Question. Find the coordinate of the point P where the line through A(3, –4, –5) and B(2, –3, 1) crosses the plane passing through three points L(2, 2, 1), M(3, 0, 1) and N(4, –1, 0). Also, nd the ratio in which P divides the line segment AB.
Answer. General equation of a plane passing through
L(2, 2, 1) is a(x – 2) + b(y – 2) + c(z – 1) = 0                              …(i)
It will pass through M(3, 0, 1) and N(4, –1, 0) if
a(3 – 2) + b(0 – 2) + c(1 – 1) = 0
⇒ a – 2b + 0 · c = 0                              …(ii)
a(4 – 2) + b(–1– 2) + c(0 – 1) = 0
⇒ 2a – 3b – c = 0                              …(iii)
Solving (ii) and (iii), we get

Question.

the equation of the plane containing these lines.

Here, x1 = –3, y1 = 1, z1 = 5;
a1 = –3, b1 = 1, c1 = 5;
x2 = –1, y2 = 2, z2 = 5;
a2 = –1, b2 = 2, c2 = 5

⇒ (x + 3) (5 – 10) – (y – 1) (–15 + 5) + (z – 5) (– 6 + 1) = 0
⇒ –5x – 15 + 10y – 10 – 5z + 25 = 0
⇒ 5x – 10y + 5z = 0 ⇒ x – 2y + z = 0

Question. The given lines are

find the value of k and hence find the equation of plane containing these lines.

⇒ (x – 1) (–20 – 2) – (y –2) (–15 – 4) + (z – 3) (–3 + 8) = 0
⇒ –22 (x – 1) + 19(y – 2) + 5(z – 3) = 0
⇒ –22x + 19y + 5z = 31.

Question. Find the vector equation of the plane which contains the line of intersection of the planes

Answer. The equation of any plane through the line of intersection of the given planes is

Question. Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 6.
Answer. Equation of plane passing through (– 1, 3, 2) is
a(x + 1) + b(y – 3) + c(z – 2) = 0                               …(i)
Since it is perpendicular to planes,
x + 2y + 3z = 5 and 3x + 3y + z = 6.
Then, a + 2b + 3c = 0                               …(ii)
and 3a + 3b + c = 0                               …(iii)
Solving (ii) and (iii), we get

⇒ a = – 7λ, b = 8λ, c = – 3λ
Putting values of a, b, c in (i), we get
– 7λ(x + 1) + 8λ(y – 3) – 3λ(z – 2) = 0
⇒ – 7x – 7 + 8y – 24 – 3z + 6 = 0
⇒ 7x – 8y + 3z + 25 = 0

Question. Find the equation of the plane passing through the point (–1, –1, 2) and perpendicular to each of the following planes 2x + 3y – 3z = 2 and 5x – 4y + z = 6.
Answer. Equation of plane passing through the point
(–1, –1, 2) is
a(x + 1) + b(y + 1) + c(z –2) = 0                               …(i)
Also, this plane is perpendicular to each of the planes
2x + 3y – 3z = 2 and 5x – 4y + z = 6
∴ 2a + 3b – 3c = 0                               …(ii)
and 5a – 4b + c = 0                             …(iii)
Solving (ii) and (iii), we get

⇒ a = –9λ, b = – 17λ, c = – 23λ
Putting the values of a, b, c in (i), we get
–9λ(x + 1) – 17λ(y + 1) – 23λ(z –2) = 0
⇒ –9x – 9 – 17y – 17 – 23z + 46 = 0
⇒ 9x + 17y + 23z – 20 = 0.

Question. Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector 2î +3ĵ + 4k̂ to the plane r̅ · (2î + ĵ + 3k̂) − 26 = 0. Also find image of P in the plane.
Answer.  Let M be the foot of the perpendicular drawn from the point P(2, 3, 4) on the plane
r̅ · (2î + ĵ + 3k̂) − 26 = 0
i.e., 2x + y + 3z – 26 = 0
Then, PM is the normal to the plane. So, the direction ratios of PM are proportional to 2, 1, 3.
So, the equation of PM is

∴ The coordinates of M ≡ (2r + 2, r + 3, 3r + 4).
Since M lies on the plane 2x + y + 3z – 26 = 0
∴ 2(2r + 2) + r + 3 + 3(3r + 4) – 26 = 0
⇒ 4r + 4 + r + 3 + 9r + 12 – 26 = 0
⇒ 14r – 7 = 0 ⇒ r = 1/2

Question. Find the distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line whose direction cosines are proportional to 2, 3, –6.
Answer. Equation of line passing through (1, –2, 3) and having direction cosines proportional to 2, 3, –6 is

Question. Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0. Then find the distance of plane thus obtained from the point A(1, 3, 6).
2x + 3y + 4z = 5                                 …(i)
and x + y + z = 1                                 …(ii)
Eq. of any plane through the line of intersection of these planes is given by
2x + 3y + 4z – 5 + λ (x + y + z – 1) = 0
⇒ (2 + λ) x + (3 + λ)y + (4 + λ)z – (5 + λ) = 0                                 …(iii)
Q This plane is ⊥ to the plane x – y + z = 0
∴ (2 + λ)·1 + (3 + λ)·(–1) + (4 + λ).1 = 0
⇒ 2 + λ – 3 – λ + 4 + λ = 0 ⇒ λ = –3
∴ Eq. of the required plane is
(2 – 3)x + (3 – 3)y + (4 – 3)z – (5 – 3) = 0
⇒ –x + z – 2 = 0 ⇒ x – z + 2 = 0
Now distance of the point (1, 3, 6) from this plane

Question. Find the distance of the point (3, 4, 5) from the plane x + y + z = 2 measured parallel to the line 2x = y = z.

Question. Find the coordinates of the foot of the perpendicular, the equation of the perpendicular and the perpendicular distance of the point P(3, 2, 1) from the plane 2x – y + z + 1 = 0. Find also the image of the point P in the plane.
Answer. The given plane is 2x – y + z + 1 = 0                                …(i)
Perpendicular distance from P(3, 2, 1) to (i)

Question. Find the coordinates of the image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0.
Answer. Let Q be the foot of the perpendicular from P (1, 3, 4) to the plane 2x – y + z + 3 = 0, then PQ is normal to the plane. So, its direction ratios are proportional to 2, –1, 1.
Since PQ passes through (1, 3, 4), therefore, its equation is

Let the coordinates of Q be (2λ + 1, –λ + 3, λ + 4)
Since Q lies on the plane 2x – y + z + 3 = 0
∴ 2(2λ + 1) – (–λ + 3) + (λ + 4) + 3 = 0
⇒ 4λ + 2 + λ – 3 + λ + 4 + 3 = 0
⇒ 6λ + 6 = 0 ⇒ λ = –1
So, the coordinates of Q are (–1, 4, 3)
Let R (x1 y1 z1) be the image of P (1, 3, 4) and as Q is the mid-point of PR,

⇒ x1 = –3, y1 = 5, z1 = 2
So, the image of the point P is R (–3, 5, 2).

Question. Find the distance of the point (–2, 3, –4) from the line

measured parallel to the plane 4x + 12y – 3z + 1 = 0.
Answer. The given point be P(–2, 3, –4) and the equation

Question. Find the vector and cartesian forms of the equation of the plane passing through the point (1, 2, –4) and parallel to the lines
r̅ = î + 2ĵ − 4k̂ + λ (2î + 3ĵ + 6k̂) and
r̅ = î − 3ĵ + 5k̂ + μ (î + ĵ − k̂).
Also, find the distance of the point (9, –8, –10) from the plane thus obtained.

Question. Find the equation of the plane passing through the line of intersection of the planes r̅ · (î + ĵ + k̂) = 1 and  r̅ · (2î + 3ĵ − k̂) + 4 = 0 and parallel to x-axis.
Answer. Here, r̅ · (î + ĵ + k̂) = 1
⇒ x + y + z – 1 = 0                                                     …(i)
and r̅ · (2î + 3ĵ − k̂) + 4 = 0
⇒ 2x + 3y – z + 4 = 0                                                     …(ii)
Eq. of the plane passing through the line of intersection of the planes (i) and (ii) is given by
x + y + z – 1 + λ(2x + 3y – z + 4) = 0
⇒ (1 + 2λ)x + (1 + 3λ)y + (1 – λ)z + (4λ – 1) = 0                       …(iii)
∵ The plane is parallel to x-axis whose d.r’s are 1, 0, 0.
∴ (1 + 2λ) · 1 + (1 + 3λ) · 0 + (1 – λ) · 0 = 0
⇒ 1+ 2 = 0 ⇒ λ = – 1/2
Substituting value of λ in (iii), we get

Question. Find the vector equation of the line passing through the point (1, 2, 3) and parallel to the planes r̅ · (î − ĵ + 2k̂) = 5 and r̅ · (3î + ĵ + k̂) = 6.
Answer. Any line through (1, 2, 3) is

Question. Find the equation of the plane through the line of intersection of the planes 2x + y – z = 3 and 5x – 3y + 4z + 9 = 0 and parallel to the line

Answer. The equation of the plane passing through the line of intersection of the planes 2x + y – z = 3 and
5x – 3y + 4z + 9 = 0 is
(2x + y – z – 3) + λ(5x – 3y + 4z + 9) = 0
⇒ x(2 + 5λ) + y(1 – 3λ) + z(4λ – 1) + 9λ – 3 = 0                                        …(i)
Since, plane (i) is parallel to the line

Question. Find the vector and cartesian equations of a plane containing the two lines
r̅ = 2î + ĵ − 3k̂ + λ(î + 2ĵ + 5k̂) and
r̅ = 3î + 3ĵ + 2k̂ + μ(3î − 2ĵ + 5k̂)
Also show that the line
r̅ = (2î + 5ĵ + 2k̂) + p(3î − 2ĵ + 5k̂) lies in the plane.

Question. Find the equation of the plane passing through the point P(1, 1, 1) and containing the line
r̅ = (−3î + ĵ + 5k̂) + λ(3î − ĵ − 5k̂).
Also, show that the plane contains the line
r̅ = (−î + 2ĵ + 5k̂) + μ (î − 2ĵ −5k̂).
Answer. The equation of plane passing through the point P(1, 1, 1) is
a(x – 1) + b(y – 1) + c(z – 1) = 0                               …(i)
Since plane contains the line,
r̅ = − 3î + ĵ + 5k̂ + λ (3î − ĵ − 5k̂)
∴ Point (–3, 1, 5) lies in the plane.
So, a(–3–1) + b(1 –1) + c(5 – 1) = 0
⇒ –4a + 4c = 0 ⇒  a – c = 0                               …(ii)
Direction ratios of plane are normal to direction ratios of line
∴ 3a – b – 5c = 0                               …(iii)
Solving (ii) and (iii), we get

⇒ a = –λ, b = 2λ, c = –λ
Substituting the values of a, b, c in (i), we get
–λ(x – 1) + 2λ(y – 1) –λ(z – 1) = 0
⇒ x – 2y + z = 0
∴ Vector equation of plane is r̅ · (î −2ĵ + k̂)=0.
This plane contains the line
r̅ = −î + 2ĵ + 5k̂ + μ(î − 2ĵ − 5k̂)
iff (î −2ĵ + k̂)· (î − 2ĵ − 5k̂) = 0
⇒ 1 + 4 – 5 = 0 which is true.
Also, the point (–1, 2, 5) should satisfy the equation of plane.
∴ 1(–1) –2(2) + 1(5) = 0
⇒ –1 – 4 + 5 = 0, which is true.
Hence, the plane r̅ (î − 2ĵ + k̂) = 0 contains the line r̅ = −î + 2ĵ + 5k̂ + μ(î − 2ĵ − 5k̂).