Question

The point of intersection of the line $\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}$ and plane $2 x-y+3 z-1= 0$ is.

• A. (10, -10, 3)
• B. (10, 10, -3)
• C. (-10, 10, 3)
• D. none of these

Answer 1

Answer: (B)

(10, 10, -3)

Answer 2

We have line, $\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}=\lambda$ ...(1)
General point $P$ on the line is,
$P =(3 \lambda+1,4 \lambda-2,-2 \lambda+3)$
since line (1) intersect plane $2 x-y+3 z-1=0$,
Assume it intersects at a point $P$
Therefore,
$2(3 \lambda+1)-(4 \lambda-2)+3(-2 \lambda+3)-1=0$
$6 \lambda+2-4 \lambda+2-6 \lambda+9-1=0$
$4 \lambda=12$
$\lambda=3$
Therefore,
$P =(10,10,-3))$