JEE Math Practice Question
The shortest distance between the lines $${{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}}$$ and $${{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}}$$ is :
- A.$$7\sqrt 3 $$
- B.$$5\sqrt 3 $$
- C.$$4\sqrt 3 $$
- D.$$6\sqrt 3 $$
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