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  4. The shortest distance between the lines (x-3/3) = (y-8/-1) = (z-3/1) and (x+3/-3) = (y+7/2) = (z-6/4) is :

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Q. The shortest distance between the lines
$\frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1}$ and $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}$ is :

JEE MainJEE Main 2020Three Dimensional Geometry

Solution:

Shortest distance $=\frac{\begin{Vmatrix}6&15&-3\\ 3&-1&1\\ -3&2&4\end{Vmatrix}}{\sqrt{11\times29-49}}=\frac{270}{\sqrt{270}}$
$=\sqrt{270}=3\sqrt{30}$