Question

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

• A. $2\sqrt{11}$
• B. $\sqrt{126}$
• C. $13$
• D. $14$

Answer 1

The correct option is C $13$

Let $\frac{x-2}{2}=\frac{y+1}{4}=\frac{z-2}{12}=k$
Then, $x=2k+2$
$y=4k-1$
$z=12k+2$
Substitute in the equation of plane
$x-y+z=5$
$10k+5=5$
$k=0$
Point of intersection $=(2,-1,2)$
Distance from $(-1,-5,-10)=$ $\sqrt{3^2+4^2+{12}^2}=13$