JEE Math Practice Question
The shortest distance between the lines $$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$$ and $$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$$ is:
- A.$$\frac{185}{\sqrt{563}}$$
- B.$$\frac{187}{\sqrt{563}}$$
- C.$$\frac{178}{\sqrt{563}}$$
- D.$$\frac{179}{\sqrt{563}}$$
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