Question

The distance of the point (1, 0, 2) from the point of intersection of the line
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$
and the plane $x-y+z=16$, is

• A. $2\sqrt{14}$
• B. 8
• C. $3\sqrt{21}$
• D. 13

Answer 1

The correct option is D 13

We will take the point of intersection as the general point on the line.
$(3\lambda +2,4\lambda -1,12\lambda +2)$
Now we will substitute this in the equation of the plane.
$3\lambda +2-4\lambda +1+12\lambda +2=16$
$11\lambda =11\Rightarrow \lambda =1$
(5, 3, 14) is the point of intersection.
Distance = $\sqrt{16+9+144}=\sqrt{169}=13$