Q.The shortest distance between the lines $\dfrac{x+1}{7} = \dfrac{y+1}{-6} = \dfrac{z+1}{1}$ and $\dfrac{x-3}{1} = \dfrac{y-5}{-2} = \dfrac{z-7}{1}$ is

\sqrt{29}

\sqrt{31}

\sqrt{33}

\sqrt{35}

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+17=y+1−6=z+11\dfrac{x+1}{7} = \dfrac{y+1}{-6} = \dfrac{z+1}{1}7x+1​=−6y+1​=1z+1​ and x−31=y−5−2=z−71\dfrac{x-3}{1} = \dfrac{y-5}{-2} = \dfrac{z-7}{1}1x−3​=−2y−5​=1z−7​ is