The shortest distance between the lines

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$

$\frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}$ is :

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

The correct answer is Option (1) → $\frac{\sqrt{293}}{7}$

$\vec{a_1}=\hat i+2\hat j-4\hat k$

$\vec{a_2}=3\hat i+3\hat j-5\hat k$

($L_2$ is parallel to as $4\hat i+6\hat j+12\hat k$ ⇒ $l_2$ parallel to $2\hat i+3\hat j+6\hat k$)

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

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

$=\frac{|(2\hat i+3\hat j+6\hat k)×(2\hat i+\hat j-\hat k)|}{7}$

$=\frac{|-9\hat i+14\hat j-4\hat k|}{7}$

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