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,
$\begin{aligned}
& \vec{a}_1=-2 \hat{j}+\hat{k}, b_1+6 \hat{i}+6 \hat{j}+\hat{k}, \vec{a}_2=-\hat{i}, \\
& \vec{b}_2=12 \hat{i}+6 \hat{j}-\hat{k} \\
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & k \\
6 & 6 & 1 \\
12 & 6 & -1
\end{array}\right|=-12 \hat{i}+18 \hat{j}-36 \hat{k} \\
& \text { Shortest distance }=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1-\vec{b}_2\right)\right|}{\left|\vec{b}_1 \times \vec{b}_2\right|}
\end{aligned}$
$\begin{aligned} & =\frac{|(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(-12 i+18 \hat{j}-36 \hat{k})|}{\sqrt{(-12)^2+(18)^2+(-36)^2}} \\ & =\frac{|+12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2\end{aligned}$