Let the plane P pass through the intersection of the planes $2 x+3 y-z=2$ and $x+2 y+3 z=6$, and be perpendicular to the plane $2 \mathrm{x}+\mathrm{y}-\mathrm{z}+1=0$. If d is the distance of P from the point $(-7,1,1)$, then $\mathrm{d}^{2}$ is equal to :

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Three Dimensional Geometry

Let the plane P pass through the intersection of the planes

2 x + 3 y - z = 2

and

x + 2 y + 3 z = 6

, and be perpendicular to the plane

2 x + y - z + 1 = 0

. If d is the distance of P from the point

(- 7, 1, 1)

, then

d^{2}

is equal to :

Answer: (A)

A plane through the intersection of the first two is

(2 + λ) x + (3 + 2 λ) y + (3 λ - 1) z - (2 + 6 λ) = 0

. Perpendicularity to

2 x + y - z + 1 = 0

requires the normals' dot product to vanish:

2 (2 + λ) + (3 + 2 λ) - (3 λ - 1) = 0

, i.e.

8 + λ = 0

, so

λ = - 8

and the plane becomes

6 x + 13 y + 25 z - 46 = 0

. Its distance from

(- 7, 1, 1)

\frac{∣ - 42 + 13 + 25 - 46 ∣}{36 + 169 + 625} = \frac{50}{830}

. Hence

d^{2} = \frac{2500}{830} = \frac{250}{83}

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option