Question:
The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2 \mathrm{x}-\mathrm{y}+\mathrm{z}=6$ from the point $(-1,-1,2)$ is______.
Solution:
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}=\lambda$
$x=2 \lambda+1, y=3 \lambda+2, z=6 \lambda-1$
for point of intersection of line \& plane
$2(2 \lambda+1)-(3 \lambda+2)+(6 \lambda-1)=6$
$7 \lambda=7 \Rightarrow \lambda=1$
point: $(3,5,5)$
$(\text { distance })^{2}=(3+1)^{2}+(5+1)^{2}+(5-2)^{2}$
$=16+36+9=61$