$\left(3,5,7\right)$ satisfy the line $L_1$

$\frac{3-a}{\ell }=\frac{5-2}{3}=\frac{7-b}{4}$

$\frac{3-a}{\ell }=1 \& \frac{7-b}{4}=1$

$a+\ell =3 \dots \left(1\right) \& b=3 ...\left(2\right)$

${\overrightarrow{v}}_1·{\overrightarrow{v}}_2=0\Rightarrow \ell -6+4=0\Rightarrow \ell =2$

$a+\ell =3\Rightarrow a=1$

$L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$

$L_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$

$\overrightarrow{p}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}$

$\overrightarrow{q}=3\hat{\mathrm{i}}+4\hat{\mathrm{j}}+5\hat{\mathrm{k}}$

$\overrightarrow{p}\times \overrightarrow{q}=-\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}$

Shortest distance $=\left|\frac{\overrightarrow{AB}·\left(\overrightarrow{p}\times \overrightarrow{q}\right)}{\left|\overrightarrow{p}\times \overrightarrow{q}\right|}\right|=\frac{1}{\sqrt{6}}$