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

Let the equation of the plane, that passes through the point

(1, 4, - 3)

and contains the line of intersection of the planes

3 x - 2 y + 4 z - 7 = 0

and

x + 5 y - 2 z + 9 = 0

, be

α x + β y + γ z + 3 = 0

, then

α + β + γ

is equal to:

(A)23
(B)15
(C)-15
(D)-23✓ Correct

Answer: (D)

Solution

Any plane through the line of intersection is

(3 x - 2 y + 4 z - 7) + λ (x + 5 y - 2 z + 9) = 0

. Substituting

(1, 4, - 3)

gives

(3 - 8 - 12 - 7) + λ (1 + 20 + 6 + 9) = 0

, i.e.

- 24 + 36 λ = 0

, so

λ = \frac{2}{3}

. The plane becomes

\frac{11}{3} x + \frac{4}{3} y + \frac{8}{3} z - 1 = 0

; multiplying by

- 3

gives

- 11 x - 4 y - 8 z + 3 = 0

, of the required form

α x + β y + γ z + 3 = 0

. Hence

α + β + γ = - 11 - 4 - 8 = - 23

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option

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