The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x2+y2+z2+4x−2y−6z=155 is
The correct option is B 13
The centre of the sphere is (-2, 1, 3) and its radius is √4+1+9+155=13.
Length of the perpendicular from the centre of the sphere on the plane is ∣∣∣−24+4+9−327√144+16+9∣∣∣=33813=26
So the plane is outside the sphere and the required distance is equal to 26 - 13 = 13.