JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Three Dimensional Geometry

The square of the distance of the point of intersection of the line

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 1}{6}

and the plane

2 x - y + z = 6

from the point

(- 1, - 1, 2)

is ____.

Answer: 61

Solution

Parametrise the line as

x = 1 + 2 λ, y = 2 + 3 λ, z = - 1 + 6 λ

. Substituting into

2 x - y + z = 6

2 (1 + 2 λ) - (2 + 3 λ) + (- 1 + 6 λ) = 6

, i.e.

7 λ - 1 = 6

, so

λ = 1

and the point is

(3, 5, 5)

. Its square distance from

(- 1, - 1, 2)

(3 + 1)^{2} + (5 + 1)^{2} + (5 - 2)^{2} = 16 + 36 + 9 = 61

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Integer answer

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