If two opposite sides of a tetrahedron of volume 12 are 3 and 4 and angle between them is $\frac{\pi }{6}$ . If shortest distance between them is  $\lambda$ then  $\sqrt{\frac{\lambda }{3}}=$

Let  $\overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{c}$ are coter minus edges
$v=\frac{1}{6}\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$
$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=72\to \left(1\right)$
Let  $\left|\overrightarrow{a}\right|=3,$
$\left|\overrightarrow{BC}\right|=4$
Equation of  $\overrightarrow{OA}$ is   $\overrightarrow{r}=(1-t)\overrightarrow{O}+t\overrightarrow{a}=\overrightarrow{O}+t\overrightarrow{a}\to \left(2\right)$
Equation of  $\overrightarrow{BC}$ is  $\overrightarrow{r}=(1-s)\overrightarrow{b}+s\overrightarrow{c}=\overrightarrow{b}+s(\overrightarrow{c}-\overrightarrow{b})\to \left(3\right)$
Shortest distance between (2) and (3) is =  $\frac{[\overrightarrow{b}\overrightarrow{c}-\overrightarrow{b}\overrightarrow{a}]}{|\overrightarrow{a}\times (\overrightarrow{c}-\overrightarrow{b})|}=\lambda$
$\therefore \lambda =\frac{\overrightarrow{b}\times (\overrightarrow{c}-\overrightarrow{b}).\overrightarrow{a}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{BC}\right|\sin \frac{\pi }{6}}=\frac{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}{\left(3\right)\left(4\right)\left(\frac{1}{2}\right)}=\frac{72}{6}=12$
$\therefore \sqrt{\frac{\lambda }{3}}=\sqrt{4}=2$