Given a tetrahedron $\text{DABC }$with $\mathrm{AB} =12, \mathrm{CD}=6.$ If the shortest distance between the skew lines $\mathrm{AB}$
and $\mathrm{CD}$is $8$and the angle between them is $\frac{\sqrt{\pi }}{6}$ then the $\frac{1}{6}$ of the volume of the tetrahedron is 

Give $|\overrightarrow{b}-\overrightarrow{a}|=12,\left|\overrightarrow{c}\right|=6$ Equation of CD is $\overline{r}=\overline{tc}$ and Equation of AB is 

$\overrightarrow{r}=\overrightarrow{a}+s(\overrightarrow{b}-\overrightarrow{a})\overrightarrow{r}=\overrightarrow{a}+s(\overrightarrow{b}-\overrightarrow{a})\overrightarrow{r}=\overrightarrow{a}+s(\overrightarrow{b}-\overrightarrow{a})$

$S.D=\frac{\left|\right[\overrightarrow{a}\overrightarrow{b}-\overrightarrow{a}\overrightarrow{c}\left]\right|}{|\overrightarrow{b}-\overrightarrow{a}|\times \overrightarrow{c}}=8\Rightarrow \left|\right[\overrightarrow{a}\overrightarrow{c}\overrightarrow{b}\left]\right|=8\left|\right(\overrightarrow{b}-\overrightarrow{a})\times \overrightarrow{c}|$

$\Rightarrow 6V=8\frac{\left|\overrightarrow{c}\right||\overrightarrow{b}-\overrightarrow{a}|}{2}$

$\Rightarrow V=\frac{8\times 6\times 12}{6\times 2}=48\Rightarrow \frac{1}{6}v=8$