Answer
$${\left( {x - 3} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {z - 6} \right)^2} = {\left( {\frac{{15}}{2}} \right)^2}$$
Work Step by Step
$$\eqalign{
& {\text{Let the center be }}\underbrace {\left( {3, - 2,6} \right)}_{\left( {{x_0},{y_0},{z_0}} \right)},{\text{ Diameter: 15}} \to {\text{radius: }}r = \frac{{15}}{2} \cr
& {\text{The standard equation of a sphere of radius }}r,{\text{ centered at}} \cr
& \left( {{x_0},{y_0},{z_0}} \right){\text{ is:}} \cr
& {\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} + {\left( {z - {z_0}} \right)^2} = {r^2} \cr
& {\text{Then,}} \cr
& {\left( {x - 3} \right)^2} + {\left( {y - \left( { - 2} \right)} \right)^2} + {\left( {z - 6} \right)^2} = {\left( {\frac{{15}}{2}} \right)^2} \cr
& {\left( {x - 3} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {z - 6} \right)^2} = {\left( {\frac{{15}}{2}} \right)^2} \cr} $$
