Question

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

• A. $10$
• B. $11$
• C. $12$
• D. $13$

Answer 1

The correct option is D $13$

Given line is $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=t$ (say)
Then, any point on this line is $(3t+2,4t-1,12t+2)$.
This point lies on the plane $x-y+z=5$.
Therefore, $3t+2-4t+1+12t+2=5$
$\Rightarrow 11t=0\Rightarrow t=0$
The point is $(2,-1,2)$
Its distance from $(-1,-5,-10)$
$=\sqrt{{(2+1)}^2+{(-1+5)}^2+{(2+10)}^2}$
$=\sqrt{9+16+144}=13$