JEE Math Practice 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 :
- A.$${{18} \over {\sqrt 5 }}$$
- B.$${{22} \over {3\sqrt 5 }}$$
- C.$${{46} \over {3\sqrt 5 }}$$
- D.$$6\sqrt 3 $$
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