The distance of the point

equation of line parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ passes

through $(1,-2,3)$ is

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

$x=2 r+1$

$y=3 r-2$

$z=-6 r+3$

So $2 r+1-3 r+2-6 r+3=5$

$\Rightarrow$ $-7 r+1=0$

$r=\frac{1}{7}$

$x=\frac{9}{7}, y=\frac{-11}{7},, z=\frac{15}{7}$

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

$=\sqrt{\left(\frac{2}{7}\right)^{2}+\left(\frac{3}{7}\right)^{2}+\left(\frac{6}{7}\right)^{2}}$

$=\frac{1}{7} \sqrt{4+9+36}$

$=\frac{1}{7} \sqrt{49}=1$