JEE Math Practice Question
The shortest distance between the lines $$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$$ and $$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$$ is
- A.$$8 \sqrt{3}$$
- B.$$6 \sqrt{3}$$
- C.$$5 \sqrt{3}$$
- D.$$4 \sqrt{3}$$
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