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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Numerical Value Type Questions - 3-Dimensional Geometry

  • question_answer
    If the shortest distance between the line \[\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}\] and \[\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}\] is \[k\sqrt{30}\] then find k.

    Answer:

    3.00


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