Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given lines are

[tex]\tt\vec{r}=(4 \hat{i}-\hat{j})+\lambda (\hat{i}+2 \hat{j}-3 \hat{k}) - - - - (1) \\ \\ [/tex]

and

[tex]\tt\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+4 \hat{j}-5 \hat{k}) - - - (2) \\ \\ [/tex]

So, we have

[tex]\tt \: a_1=4 \hat{i}-\hat{j} \\ [/tex]

[tex]\tt \: a_2=\hat{i}-\hat{j} + 2\hat{k} \\ [/tex]

[tex]\tt \: b_1=\hat{i} + 2\hat{j} - 3\hat{k} \\ [/tex]

[tex]\tt \: b_2=2\hat{i} + 4\hat{j} - 5\hat{k} \\ [/tex]

Now, Consider

[tex] \red{ \tt \: a_2 - a_1 \: } \\ [/tex]

[tex] \tt \: = \: (\hat{i} - \hat{j} + 2\hat{k}) - (4\hat{i} - \hat{j}) \\ [/tex]

[tex] \tt \: = \: \hat{i} - \hat{j} + 2\hat{k} - 4\hat{i} + \hat{j}\\ [/tex]

[tex] \tt \: = \: - 3\hat{i} + 2\hat{k}\\ \\ [/tex]

Now, Consider

[tex] \red{ \tt \:b_1 \times b_2 }\\ [/tex]

[tex]\rm \: = \: \begin{gathered}\sf \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\1&2& - 3\\2& 4& - 5\end{array}\right | \end{gathered} \\ [/tex]

[tex] \tt \: = \: ( - 10 + 12)\hat{i} - \hat{j}( - 5 + 6) + \hat{k}(4 - 4) \\ [/tex]

[tex] \tt \: = \: 2\hat{i} -\hat{j} \\ \\ [/tex]

Now, Consider

[tex] \red{ \tt \: |b_1 \times b_2| } \\ [/tex]

[tex] \tt \: = \: \sqrt{ {(2)}^{2} + {( - 1)}^{2} } \\ [/tex]

[tex] \tt \: = \: \sqrt{ 4 + 1} \\ [/tex]

[tex] \tt \: = \: \sqrt{5} \\ \\ [/tex]

Now, Consider

[tex] \red{ \tt \: (a_2 - a_1).(b_1 \times b_2)} \\ [/tex]

[tex] \tt \: = \: ( - 3\hat{i} + 2\hat{k}).(2\hat{i} - \hat{j})[/tex]

[tex] \tt \: = \: - 6 \\ \\ [/tex]

Now, Minimum distance between two lines is given by

[tex] \red{ \tt \: = \: \dfrac{ |(a_2 - a_1).(b_1 \times b_2)| }{ |b_1 \times b_2| } } \\ [/tex]

[tex] \tt \: = \: \dfrac{ | - 6| }{ \sqrt{5} } \\ [/tex]

[tex] \tt \: = \: \dfrac{6}{ \sqrt{5} } \\ [/tex]

[tex] \tt \: = \: \dfrac{6 \sqrt{5} }{ 5 } \: units \\ \\ [/tex]