Q.The shortest distance between the lines $\dfrac{x-3}{3} = \dfrac{y-8}{ -1} = \dfrac{z-3}{1}$ and $\dfrac{x+3}{ -3} = \dfrac{y+7}{2} = \dfrac{z-6}{4}$ is

\sqrt{3}

2\sqrt{3}

3\sqrt{3}

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.The shortest distance between the lines x−33=y−8−1=z−31\dfrac{x-3}{3} = \dfrac{y-8}{ -1} = \dfrac{z-3}{1}3x−3​=−1y−8​=1z−3​ and x+3−3=y+72=z−64\dfrac{x+3}{ -3} = \dfrac{y+7}{2} = \dfrac{z-6}{4}−3x+3​=2y+7​=4z−6​ is