JEE Math Practice Question
The shortest distance between the lines $${{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5}$$ and $${{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}}$$ is :
- A.$$2\sqrt3$$
- B.$$3\sqrt3$$
- C.$$4\sqrt3$$
- D.$$5\sqrt3$$
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