The lines are $\frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}$ and $\frac{x+1}{12}=\frac{y}{6}=\frac{z}{-1}$
Here,
$$
\begin{array}{l}
\vec{a}_{1}=-2 \hat{j}+\hat{k}, b_{1}+6 \hat{i}+6 \hat{j}+\hat{k}, \vec{a}_{2}=-\hat{i} \\
\vec{b}_{2}=12 \hat{i}+6 \hat{j}-\hat{k} \\
\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & k \\
6 & 6 & 1 \\
12 & 6 & -1
\end{array}\right|=-12 \hat{i}+18 \hat{j}-36 \hat{k}
\end{array}
$$
$$
\begin{array}{l}
\text { Shortest distance }=\frac{\left|\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1}-\vec{b}_{2}\right)\right|}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|} \\
=\frac{|(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(-12 i+18 \hat{j}-36 \hat{k})|}{\sqrt{(-12)^{2}+(18)^{2}+(-36)^{2}}} \\
=\frac{|+12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2
\end{array}
$$