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 $2 x-y+z=6$ from the point $(-1,-1,2)$ is

Equation of line

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$

Let $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}=\lambda$

$\Rightarrow x = 2\lambda +1, y = 3\lambda +2, z=6\lambda -1$

$\Rightarrow$ Let $P(2\lambda +1,3\lambda +2,6\lambda -1)$ is point of intersection of line and plane

So it should satisfy the given plane.

$\Rightarrow 2(2\lambda +1)-3\lambda -2+6\lambda -1=6$

$7\lambda =7\Rightarrow \lambda =1$

Point $\mathrm{P}(3,5,5)$

Square of distance from point $(-1,-1,2)$  and $\mathrm{P}(3,5,5)$

$=\sqrt{{\left(x_2-x_1\right)}^2 +{\left(y_2-y_1\right)}^2 +{\left(z_2-z_1\right)}^2 }$

$=4^2+6^2+3^2=61$