The shortest distance between the lines $I_1$ and $I_2$ given by

$I_1: \vec{r}=\hat{i} + 2\hat{j}-4\hat{k}+ \lambda (2\hat{i} + 3\hat{j} + 6\hat{k} ) $ and $I_2 : \vec{r}=3\hat{i}+3\hat{j}-5\hat{k} + \mu (2\hat{i} + 3\hat{j}+6\hat{k})$ is :

$\frac{\sqrt{293}}{7}$

The correct answer is Option (3) → $\frac{\sqrt{293}}{7}$

Given lines are parallel

$\vec{a_1}=\hat i+2\hat j-4\hat k$, $\vec{a_2}=3\hat i+3\hat j-5\hat k$

$\vec b=2\hat i+3\hat j+6\hat k$

so shortest distance = $\frac{\left|(\vec{a_2}-\vec{a_1})×\vec b\right|}{|\vec b|}$

$=\frac{\left|(2\hat i+\hat j-\hat k)×(2\hat i+3\hat j+6\hat k)\right|}{\sqrt{2^2+3^2+6^2}}$

$=\frac{|9\hat i-14\hat j+4\hat k|}{\sqrt{49}}$

$=\frac{\sqrt{293}}{7}$