The shortest distance from the plane $$12x+4y+3z=327$$ to the sphere $${x^2} + {y^2} + {z^2} + 4x - 2y - 6z = 155$$ is
JEE · Math · previous-year question
- A.$$39$$
- B.$$26$$
- C.$$11{4 \over {13}}$$
- D.$$13$$correct
Answer
D. $$13$$
Explanation
Shortest distance $$=$$ perpendicular distance between the plane and sphere $$=$$ distance of plane from center of sphere $$-$$ radius $$ = \left| {{{ - 2 \times 12 + 4 \times 1 + 3 \times 3 - 327} \over {\sqrt {144 + 9 + 16} }}} \right| - \sqrt {4 + 1 + 9 + 155} $$ $$ = 26 - 13 = 13$$
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