The shortest distance from the plane 12x + y + 3z = 327 to the sphere x2+y2+z2+4x−2y−6z=155 is

The given sphere is  x2+y2+z2+4x−2y−6z=155Its centre is (-2, 1, 3) and radius=4+1+9+155=169=13Therefore, distance of centre (1, 1, 3) from the plane 12x + 4y + 3z =327 is |12(−2)+4(1)+3(3)−327|144+16+9=26Hence, the shortest distance is 13