Let the equation of the plane P containing the line $\mathrm{x}+10=\frac{8-\mathrm{y}}{2}=\mathrm{z}$ be $\mathrm{ax}+\mathrm{by}+3 \mathrm{z}=2(\mathrm{a}+\mathrm{b})$ and the distance of the plane P from the point $(1,27,7)$ be c. Then $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}$ is equal to ______.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Three Dimensional Geometry

Let the equation of the plane P containing the line

x + 10 = \frac{8 - y}{2} = z

ax + by + 3 z = 2 (a + b)

and the distance of the plane P from the point

(1, 27, 7)

be c. Then

a^{2} + b^{2} + c^{2}

is equal to ______.

Answer: 355

Solution

The line passes through

(- 10, 8, 0)

with direction

(1, - 2, 1)

. Lying in the plane, the point gives

- 10 a + 8 b = 2 (a + b)

, so

b = 2 a

; and the direction being perpendicular to the normal

(a, b, 3)

gives

a - 2 b + 3 = 0

. These yield

a = 1, b = 2

, so the plane is

x + 2 y + 3 z = 6

. The distance from

(1, 27, 7)

c = \frac{∣1 + 54 + 21 - 6∣}{1 + 4 + 9} = \frac{70}{14} = 514

. Hence

a^{2} + b^{2} + c^{2} = 1 + 4 + 350 = 355

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer

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