JEE Math Practice Question
The shortest distance between the lines $${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$$ and $${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$$ is :
- A.3
- B.$${7 \over 2}\sqrt {30} $$
- C.$$3\sqrt {30} $$
- D.$$2\sqrt {30} $$
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