The shortest distance between the lines $\frac{x+3}{1}=\frac{y-2}{2}=\frac{z+4}{3}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is:

$\frac{197}{\sqrt{237}}$

The correct answer is Option (4) - $\frac{197}{\sqrt{237}}$

$\vec{a_1}=-3\hat i+2\hat j-4\hat k,\vec{a_2}=-3\hat i-7\hat j+6\hat k$

$\vec{b_1}=\hat i+2\hat j+3\hat k,\vec{b_2}=-3\hat i+2\hat j+4\hat k$

so shortest distance

$=\left|(\vec{a_2}-\vec{a_1}).\frac{(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}\right|$

$(-9\hat j+10\hat k).\frac{2\hat i-13\hat j+8\hat k}{\sqrt{2^2+13^2+8^2}}$

$=\frac{197}{\sqrt{237}}$