The distance of the point (1,-2,3) from

Equation of line through point $P(1,-2,3)$ and parallel

to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is

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

So, any point on line $=Q(2 \lambda+1,3 \lambda-2,-6 \lambda+3)$

Since, this point lies on plane $x-y+2=5$

$\therefore 2 \lambda+1-3 \lambda+2-6 \lambda+3=5 \Rightarrow \lambda=\frac{1}{7}$

$\therefore$ Point of intersection line and plane, $Q=\left(\frac{9}{7}, \frac{11}{7}, \frac{15}{7}\right)$

$\therefore$ Required distance $P Q$

$=\sqrt{\left(\frac{9}{7}-1\right)^{2}+\left(-\frac{11}{7}+2\right)^{2}+\left(\frac{15}{7}-3\right)^{2}}=1$