Question

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

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

Answer 1

The correct option is B $(10,10,-3)$

We have line,

$\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}=\lambda ............\left(1\right)$

General point P on the line is,

$P=(3\lambda +1,4\lambda -2,-2\lambda +3)$

since line (1) intersect plane $2x-y+3z-1=0$,

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 ,

$P=(10,10,-3))$

Option (B) is correct.