JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Three Dimensional Geometry

Let the plane passing through the point

(- 1, 0, - 2)

and perpendicular to each of the planes

2 x + y - z = 2

and

x - y - z = 3

a x + b y + cz + 8 = 0

. Then the value of

a + b + c

is equal to :

(A)4✓ Correct
(B)8
(C)5
(D)3

Answer: (A)

Solution

The required normal is perpendicular to the normals

(2, 1, - 1)

and

(1, - 1, - 1)

of the two given planes, so

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

The plane through

(- 1, 0, - 2)

with this normal is

- 2 (x + 1) + (y - 0) - 3 (z + 2) = 0 \Rightarrow - 2 x + y - 3 z - 8 = 0

Multiplying by

- 1

to bring the constant term to

+ 8

gives

2 x - y + 3 z + 8 = 0

, so

a = 2

b = - 1

c = 3

and

a + b + c = 4

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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