Question

The line $\frac{x + 3}{3} = \frac{y - 2}{- 2} = \frac{z + 1}{1}$ and the plane

$4x + 5y + 3z - 5 = 0$ intersect at a point

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

Answer 1

Answer: (B)

(3, – 2, 1)

Answer 2

Line is $\frac { x + 3 } { 3 } = \frac { y - 2 } { - 2 } = \frac { z + 1 } { 1 } = \lambda$ (Let)

$x = 3 \lambda - 3 ; y = - 2 \lambda + 2 ; z = \lambda - 1$line intersects plane,

therefore, $4 ( 3 \lambda - 3 ) + 5 ( - 2 \lambda + 2 ) + 3 ( \lambda - 1 ) - 5 = 0$ $\Rightarrow \lambda = 2$.

So, $x = 3 ; y = - 2 ; z = 1$.

Trick : Since the point (3, – 2, 1) satisfies both the equations.