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 $13$

A general point on the line is $(3\lambda +2,4\lambda -1,12\lambda +2)$

Substituing this in the planes's equation,
$3\lambda +2-4\lambda +1+12\lambda +2=16\Rightarrow 11\lambda =11\Rightarrow \lambda =1$

(5, 3, 14) is the point of intersection. Distance between (5,3,14) and (1,0,2) is
$d=\sqrt{(5-1{)}^2+(3-0{)}^2+(14-2{)}^2}=\sqrt{16+9+144}=\sqrt{169}=13$