JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Three Dimensional Geometry

Let the plane containing the line of intersection of the planes
P1:

x + (λ + 4) y + z = 1

and
P2:

2 x + y + z = 2

pass through the points

(0, 1, 0)

and

(1, 0, 1)

. Then the distance of the point

(2 λ, λ, - λ)

from the plane P2 is

(A) $56$
(B) $46$
(C) $26$
(D) $36$ ✓ Correct

Answer: (D)

Solution

Take the pencil of planes

(x + (λ + 4) y + z - 1) + k (2 x + y + z - 2) = 0

through the line of intersection. Passing through

(1, 0, 1)

gives

1 + k = 0

, so

k = - 1

. Passing through

(0, 1, 0)

gives

(λ + 3) - k = 0

, so with

k = - 1

λ = - 4

. The point

(2 λ, λ, - λ)

is then

(- 8, - 4, 4)

, and its distance from

P_{2} : 2 x + y + z = 2

\frac{∣2 ( - 8 ) + ( - 4 ) + 4 - 2∣}{4 + 1 + 1} = \frac{18}{6} = 36

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Single correct option

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