A plane P contains the line of intersection of the plane \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6\) and \(\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5\). If P passes through the point \((0,2,-2)\), then the square of distance of the point (12, 12, 18) from the plane P is

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Three Dimensional Geometry

A plane P contains the line of intersection of the plane

r \cdot (\hat{i} + \hat{j} + \hat{k}) = 6

and

r \cdot (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = - 5

. If P passes through the point

(0, 2, - 2)

, then the square of distance of the point (12, 12, 18) from the plane P is

Answer: (B)

The family of planes containing the line of intersection of two given planes is

(x + y + z - 6) + λ (2 x + 3 y + 4 z + 5) = 0

. Since P passes through

(0, 2, - 2)

, substitute:

(2 - 2 - 6) + λ (6 - 8 + 5) = 0

gives

- 6 + 3 λ = 0

, so

λ = 2

. The equation of plane P is

(x + y + z - 6) + 2 (2 x + 3 y + 4 z + 5) = 0

, which simplifies to

5 x + 7 y + 9 z + 4 = 0

. The distance from

(12, 12, 18)

d = \frac{∣5 ( 12 ) + 7 ( 12 ) + 9 ( 18 ) + 4∣}{5 ^{2} + 7 ^{2} + 9 ^{2}} = \frac{310}{155}

. Therefore

d^{2} = \frac{31 0 ^{2}}{155} = 620

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option