JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Three Dimensional Geometry

The plane through the intersection of the plane

x + y + z = 1

and

2 x + 3 y - z + 4 = 0

and parallel to

y

-axis also pass through the point:

(A) $(- 3, 0, - 1)$
(B) $(- 3, 1, 1)$
(C) $(3, 3, - 1)$
(D) $(3, 2, 1)$ ✓ Correct

Answer: (D)

Solution

The planes through the given intersection form the family

(x + y + z - 1) + λ (2 x + 3 y - z + 4) = 0

, that is

(1 + 2 λ) x + (1 + 3 λ) y + (1 - λ) z + (4 λ - 1) = 0

. Being parallel to the

y

-axis kills the

y

-coefficient, so

1 + 3 λ = 0

, giving

λ = - \frac{1}{3}

and the plane

x + 4 z - 7 = 0

. Testing the candidate points, only

(3, 2, 1)

satisfies it since

3 + 4 (1) - 7 = 0

, whereas

(- 3, 0, - 1)

gives

- 3 - 4 - 7 = - 14 \neq = 0

. The point lying on the plane is

(3, 2, 1)

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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