Question

The distance of the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and me plane $x-y+z=5$ from the point $(-1,-5,-10)$ is

• A. 13
• B. 12
• C. 11
• D. 8
• E. 7

Answer 1

Answer: (A)

Given line is $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=k(say)$ Any point on the line is $(3k+2,4k-1,12k+2)$ This point lies on the plane $x-y+z=5$
$\therefore$ $3k+2-(4k-1)+12k+2=5$
$\Rightarrow$ $11k=0$
$\Rightarrow$ $k=0$
$\therefore$ Intersection point is $(2,-1,2)$ .
$\therefore$ Distance, between points $(2,-1,2)$ and $(-1,-5,-10)$
$=\sqrt{{(-1-2)}^2+{(-5+1)}^2+{(-10-2)}^2}$
$=\sqrt{9+16+144}=13$