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
$\frac{7}{2} \sqrt{30}$
$3 \sqrt{30}$
3
$2 \sqrt{30}$
Correct Option: , 2
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