A plane E is perpendicular to the two planes \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=0\) and \(\mathrm{x}-\mathrm{y}+2 \mathrm{z}=4\), and passes through the point \(\mathrm{P}(1,-1,1)\). If the distance of the plane E from the point \(\mathrm{Q}(\mathrm{a}, \mathrm{a}, 2)\) is \(3 \sqrt{2}\), then \((\mathrm{PQ})^{2}\) is equal to

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Three Dimensional Geometry

A plane E is perpendicular to the two planes

2 x - 2 y + z = 0

and

x - y + 2 z = 4

, and passes through the point

P (1, - 1, 1)

. If the distance of the plane E from the point

Q (a, a, 2)

32

, then

(PQ)^{2}

is equal to

Answer: (C)

The normal to plane E is perpendicular to both

n_{1} = (2, - 2, 1)

and

n_{2} = (1, - 1, 2)

. So the normal is

n = n_{1} \times n_{2} = (- 3, - 3, 0)

, or simplified

(1, 1, 0)

.

Plane E:

x + y + d = 0

. Since it passes through P(1, -1, 1):

1 - 1 + d = 0

, so

d = 0

. Plane E is

x + y = 0

.

Distance from Q(a, a, 2) to plane E:

\frac{∣ a + a ∣}{2} = \frac{2∣ a ∣}{2} = ∣ a ∣ 2 = 32

, so

∣ a ∣ = 3

, giving

a = \pm 3

.

For

a = 3

P Q^{2} = (3 - 1)^{2} + (3 - (- 1))^{2} + (2 - 1)^{2} = 4 + 16 + 1 = 21

.

For

a = - 3

P Q^{2} = (- 3 - 1)^{2} + (- 3 + 1)^{2} + 1 = 16 + 4 + 1 = 21

.

Thus

P Q^{2} = 21

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option