If the foot of the perpendicular from point on the line $\left(4,3,8\right)$ on the line

  then the shortest distance between the line  and the line  is equal to 

$$\begin{gathered} L_1:\frac{x-a}{2}=\frac{y-2}{m}=\frac{z-b}{4} \\ ,A(4,3,8),B(3,5,7) \end{gathered}$$ 
$$\begin{gathered} \text{drs's of }AB\text{ are }(-1,2,-1) \\ {L}_1\text{is perpendicular to }AB\Rightarrow -2+2m-4=0 \end{gathered}$$ 
$$\begin{gathered} \Rightarrow m=3 \\ B(3,5,7)\text{ lies on }{L}_1\Rightarrow a=1,b=3 \end{gathered}$$ 

$\begin{array}{r}{L}_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\\ L_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\end{array}$
$$\begin{gathered} \text{where }\overline{a}=(1,2,3),\overline{b}=(2,3,4)\overline{c}=(2,4,5),\overline{d}=(3,4,5) \\ \text{S}.\text{D}=\frac{\left|\right[\overline{a}-\overline{c} \overline{b} \overline{d}\left]\right|}{|\overline{b}\times \overline{d}|}=\frac{1}{\sqrt{6}} \end{gathered}$$