Let the equation of the plane passing through the line $x-2 y-z-5=0=x+y+3 z-5$ and parallel to the line $\mathrm{x}+\mathrm{y}+2 \mathrm{z}-7=0=2 \mathrm{x}+3 \mathrm{y}+\mathrm{z}-2$ be $\mathrm{ax}+\mathrm{by}+\mathrm{cz}=65$. Then the distance of the point $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ from the plane $2 \mathrm{x}+2 \mathrm{y}-\mathrm{z}+16=0$ is ______.

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Three Dimensional Geometry

Let the equation of the plane passing through the line

x - 2 y - z - 5 = 0 = x + y + 3 z - 5

and parallel to the line

x + y + 2 z - 7 = 0 = 2 x + 3 y + z - 2

ax + by + cz = 65

. Then the distance of the point

(a, b, c)

from the plane

2 x + 2 y - z + 16 = 0

is ______.

Answer: 9

Solution

The pencil of planes through the given line is

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

. The second line has direction

(1, 1, 2) \times (2, 3, 1) = (- 5, 3, 1)

, and making the plane's normal perpendicular to it gives

- 5 (1 + λ) + 3 (λ - 2) + (3 λ - 1) = 0

, so

λ = 12

. The plane is

13 x + 10 y + 35 z = 65

, hence

(a, b, c) = (13, 10, 35)

, whose distance from

2 x + 2 y - z + 16 = 0

\frac{∣26 + 20 - 35 + 16∣}{9} = \frac{27}{3} = 9

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Integer answer

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