Question

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

• A. 10
• B. 8
• C. 21
• D. 13

Answer 1

Answer: (D)

13

Any point on the lint $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=1$ is $(3t+2,4t-1,12t+2)$
This lies on $x-y+z=5$
$\therefore 3t+2-4t+1+12t+2=5\Rightarrow 11t=0\Rightarrow t=0$
$\therefore$ Point is $(2,-1,2)$
Its distance from $(-1,-5,-10)$ is
$=\sqrt{(2+1{)}^2+(-1+5{)}^2+(2+10{)}^2}$
$=\sqrt{9+16+144}=13$