Question

The plane through the intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to y-axis also passes through the point :

• A. (-3, 0, -1)
• B. (3, 3, -1)
• C. (3, 2, 1)
• D. (-3, 1, 1)

Answer 1

Answer: (C)

(3, 2, 1)

Answer 2

Equation of plane $(x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0$ $\Rightarrow \; (1 + 2 \lambda)x + (1 + 3 \lambda )y + (1 - \lambda)z - 1 +4\lambda = 0$ dr's of normal of the plane are $1 + 2\lambda, 1+ 3 \lambda, 1 - \lambda$ Since plane is parallel to y - axis, $1 + 3\lambda = 0$ $\Rightarrow \; \lambda = -1/3$ So the equation of plane is $x + 4z - 7 = 0$ Point $(3, 2, 1)$ satisfies this equation