Question

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

• A. $13$
• B. $1$
• C. $2$
• D. $\sqrt{13}$

Answer 1

The correct option is A $13$

Given line is $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$
Any point on the line is
$(3\lambda +2,4\lambda -1,12\lambda +2)$
On substituting in the equation of the plane, we get
$3\lambda +2-(4\lambda -1)+(12\lambda +2)=5\Rightarrow \lambda =0$
$\therefore$ Point of intersection $=(2,-1,2)$
$\therefore$ The required distance
$=\sqrt{(2+1{)}^2+(-1+5{)}^2+(2+10{)}^2}=13$