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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Self Evaluation Test - Three Dimensional Geometry

  • question_answer
    The shortest distance from the plane\[12x+4y+3z=327\] To the sphere\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+4x-2y-6z=155\] is

    A) 39

    B) 26

    C) \[11\frac{4}{13}\]

    D) 13

    Correct Answer: D

    Solution :

    [d] Shortest distance = perpendicular distance between the plane and sphere = distance of plane from centre of sphere - radius \[=\left| \frac{-2\times 12+4\times 1+3\times 3-327}{\sqrt{144+9+16}} \right|-\sqrt{4+1+9+155}\]\[=26-13=13\]


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