Q99 / 225 · JEE Main 2003 · Mathematics

The shortest distance from the plane $12 x+4 y+3 z=327$ to the sphere $x^2+y^2+z^2+4 x-2 y-6 z=155$ is

A
39
B
26
C
$11 \frac{4}{13}$
Correct answer
D
13

Solution

Shortest distance $=$ perpendicular distance $=\left|\frac{-2 \times 12+4 \times 1+3 \times 3-327}{\sqrt{144+9+16}}\right|=26$ $\therefore$ Shortest distance $$=26-\sqrt{4+1+15+9}=26-13=13 \quad[\because 26-r]$$

Hence, the correct answer is option: 'D'.

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