A line having direction ratios $⟨1,0,3⟩$cuts  planes $x-y+6z=6$ and $x-y+6z=4$ at $P$ and $Q$then $PQ.$

Suppose that  $\theta$ be the acute angle between the line whose direction ratios are $⟨1,0,3⟩$ and the plane $x-y+6z=6$

hence,  

$\begin{array}{c}\sin \theta =\frac{al+bm+cn}{\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}}\\ =\frac{1+18}{\sqrt{1+9}\sqrt{1+1+36}}\\ =\frac{19}{\sqrt{10}\sqrt{38}}\end{array}$

and $\sin \theta =\frac{PM}{PQ}$ where $PM$ is perpendicular distance from the point to the plane, or it is equal to the distance between two parallel planes. 

hence, 

$\begin{array}{c}PM=\frac{c_2-c_1}{\sqrt{a^2+b^2}}\\ =\frac{2}{\sqrt{1+1+36}}\\ =\frac{2}{\sqrt{38}}\end{array}$

Therefore, 

$\begin{array}{c}PQ=\frac{PM}{\sin \theta }\\ =\frac{2}{\sqrt{38}}\frac{\sqrt{10}\sqrt{38}}{19}\\ =\frac{2\sqrt{10}}{19}\end{array}$