Question

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. $11\frac{3}{4}$
• B. $13$
• C. $39$
• D. $26$

Answer 1

The correct option is B $13$

The centre of the sphere is (-2, 1, 3) and its radius is $\sqrt{4+1+9+155}=13$.
Length of the perpendicular from the centre of the sphere on the plane is $\left|\frac{-24+4+9-327}{\sqrt{144+16+9}}\right|=\frac{338}{13}=26$
So the plane is outside the sphere and the required distance is equal to 26 - 13 = 13.