Let P be the plane, which contains the line of intersection of the planes, $\mathrm{x}+\mathrm{y}+\mathrm{z}-6=0$ and $2 x+3 y+z+5=0$ and it is perpendicular to the $xy$-plane. Then the distance of the point $(0,0,256)$ from $P$ is equal to:

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Three Dimensional Geometry

Let P be the plane, which contains the line of intersection of the planes,

x + y + z - 6 = 0

and

2 x + 3 y + z + 5 = 0

and it is perpendicular to the

x y

-plane. Then the distance of the point

(0, 0, 256)

from

P

is equal to:

Answer: (D)

Any plane through the line of intersection can be written

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

. For it to stand perpendicular to the

x y

-plane its normal must have no

z

-component, so the coefficient of

z

, namely

1 + λ

, must vanish, giving

λ = - 1

. The plane becomes

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

, that is

- x - 2 y - 11 = 0

, or

x + 2 y + 11 = 0

. Its distance from

(0, 0, 256)

\frac{∣0 + 0 + 11∣}{1 ^{2} + 2 ^{2}} = \frac{11}{5}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option