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