The shortest distance between the lines ${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$ and ${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$, is :

Question

The shortest distance between the lines

${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$ and ${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$, is :

${{18} \over {\sqrt 5 }}$

${{22} \over {3\sqrt 5 }}$

${{46} \over {3\sqrt 5 }}$

$6\sqrt 3 $

Solution

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