Solution:

It is known to us that the shortest distance between lines with vector equations $\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$ and $\vec{r}=\overrightarrow{a_{2}}+\lambda \overrightarrow{b_{2}}$ is given as
$\left|\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|$

Given:
$\overrightarrow{\mathrm{r}}=(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda(1 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{2})$
Let’s compare it with $\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$, we get $\overrightarrow{\mathrm{a}_{1}}=(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})_{\text {and }} \overrightarrow{\mathrm{b}_{1}}=(1 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{2})$
In the similar way,
$\overrightarrow{\mathrm{r}}=(-4 \hat{\mathrm{i}}-\hat{\mathrm{k}})+\mu(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})$

Let’s compare it with $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{2}}+\lambda \overrightarrow{\mathrm{b}_{2}}$, we obtain $\overrightarrow{\mathrm{a}_{2}}=(-4 \hat{\mathrm{i}}-\hat{\mathrm{k}})_{\text {and }} \overrightarrow{\mathrm{b}_{2}}=(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{2})$

Now,
$\begin{aligned}
\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) =(-4 \hat{i}-\hat{k})-(6 \hat{i}+2 \hat{j}+2 \hat{k}) \\
=((-4-6) \hat{i}+(0-2) \hat{j}+(-1-2) \hat{k}) \\
=(-10 \hat{i}-2 \hat{j}-3 \hat{k})
\end{aligned}$

And,
$\begin{array}{l}
\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 2 \\
3 & -2 & -2
\end{array}\right| \\
\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 2 \\
3 & -2 & -2
\end{array}\right| \\
=\hat{i}[(-2 \times-2)-(-2 \times 2)]-\hat{j}[(1 \times-2)-(3 \times 2)]+\hat{k}[(1 \times-2)-(3 \times-2)] \\
=\hat{i}[4+4]-\hat{j}[-2-6]+\hat{k}[-2+6]
\end{array}$

$=8 \hat{i}+8 \hat{j}+4 \hat{k}$ So, Magnitude of $\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right| \begin{array}{l}=\sqrt{8^{2}+8^{2}+4^{2}} \\ =\sqrt{144}\end{array}=\sqrt{64+64+16}$ $=$
$=12$

Also,
$\begin{aligned}
\left(\vec{b}_{1} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) =(8 \hat{i}+8 \hat{j}+4 \hat{k}) \cdot(-10 \hat{i}-2 \hat{j}-3 \hat{k}) \\
=-80+(-16)+(-12) \\
=-108
\end{aligned}$

Therefore the shortest distance is given as
$\begin{aligned}
=\left|\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|=\left|\frac{-108}{12}\right| =|-9| \\
=9
\end{aligned}$
As a result, the shortest distance between the given two lines is 9 .