Solution:
Consider the given equations
$\begin{array}{l}
\Rightarrow \vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k} \\
\vec{r}=\hat{i}-t \hat{i}+t \hat{j}-2 \hat{j}+3 \hat{k}-2 t \hat{k} \\
\vec{r}=\hat{i}-2 \hat{j}+3 \hat{k}+t(-\hat{i}+\hat{j}-2 \hat{k}) \\
\Rightarrow \vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k} \\
\vec{r}=s \hat{i}+\hat{i}+2 \hat{j}-\hat{j}-2 \hat{k}-\hat{k} \\
\vec{r}=\hat{i}-\hat{j}-\hat{k}+s(\hat{i}+2 \hat{j}-2 \hat{k})
\end{array}$
We now need to find the shortest distance between $\vec{r}=\hat{i}-2 \hat{j}+3 \hat{k}+t(-\hat{i}+\hat{j}-2 \hat{k})$ and $\vec{r}=\hat{i}-\hat{j}-\hat{k}+s(\hat{i}+2 \hat{j}-2 \hat{k})$
It is known that shortest distance between two lines $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{1}}+\lambda \overrightarrow{\mathrm{b}_{1}}$ and $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{2}}+\mu \overrightarrow{\mathrm{b}_{2}}$ is given as:
$\mathrm{d}=\left|\frac{\left(\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}\right)}{\left|\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right|}\right|$
Oncomparing the equations we obtain, $\overrightarrow{a_{1}}=\hat{i}-2 \hat{j}+3 \hat{k}, \overrightarrow{b_{1}}=-\hat{i}+\hat{j}-2 \hat{k}$
Since,
$\left(x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{k}\right)-\left(x_{2} \hat{i}+y_{2} \hat{j}+z_{2} \hat{k}\right)=\left(x_{1}-x_{2}\right) \hat{i}+\left(y_{1}-y_{2}\right) \hat{j}+\left(z_{1}-z_{2}\right) \hat{k}$
Therefore,
$\overrightarrow{a_{2}}-\overrightarrow{a_{1}}=(\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2 \hat{j}+3 \hat{k})=\hat{j}-4 \hat{k}$
And,
$\begin{array}{l}
\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=(-\hat{i}+\hat{j}-2 \hat{k}) \times(\hat{i}+2 \hat{j}-2 \hat{k}) \\
=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 1 & -2 \\
1 & 2 & -2
\end{array}\right| \\
& =2 \hat{i}-4 \hat{j}-3 \hat{k} \\
\Rightarrow \overrightarrow{b_{1}} \times & \overrightarrow{b_{2}}=2 \hat{i}-4 \hat{j}-3 \hat{k}_{1} \quad \ldots \ldots \ldots(3) \\
\Rightarrow\left|\vec{b}_{1} \times \overrightarrow{b_{2}}\right|=\sqrt{2^{2}+(-4)^{2}+(-3)^{2}}=\sqrt{4+16+9}=\sqrt{29}
\end{array}$
Multiplying eq. (2) and eq. (3) we get,
$\begin{array}{l}
\left(\mathrm{a}_{1} \hat{\mathrm{i}}+\mathrm{b}_{1} \hat{\mathrm{j}}+\mathrm{c}_{1} \hat{\mathrm{k}}\right) \cdot\left(\mathrm{a}_{2} \hat{\mathrm{i}}+\mathrm{b}_{2} \hat{\mathrm{j}}+\mathrm{c}_{2} \hat{\mathrm{k}}\right)=\mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2} \\
\left(\overline{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right) \cdot\left(\overline{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}\right)=(2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{j}}-4 \hat{\mathrm{k}})=-4+12=8
\end{array}$
On substituting all the values in eq. (1), we get
The shortest distance between the two lines,
$\mathrm{d}=\left|\frac{8}{\sqrt{29}}\right|=\frac{8}{\sqrt{29}}$
As a result, the shortest distance is $8 \sqrt{29}$