Answer
$${\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {z - 2} \right)^2} = 17$$
Work Step by Step
$$\eqalign{
& {\text{The endpoints of the diameter are: }}\left( {0,0,4} \right),\left( {4,6,0} \right).{\text{ The center}} \cr
& {\text{is the midpoint }}\left( {\frac{{0 + 4}}{2},\frac{{0 + 6}}{2},\frac{{4 + 0}}{2}} \right) \cr
& {\text{center}} = \left( {2,3,2} \right) \cr
& {\text{The diameter is}} \cr
& d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \cr
& d = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( {6 - 0} \right)}^2} + {{\left( {0 - 4} \right)}^2}} \cr
& d = \sqrt {68} \cr
& r = \frac{d}{2} = \frac{{\sqrt {68} }}{2} = \sqrt {17} \cr
& {\text{Let the center be }}\underbrace {\left( {2,3,2} \right)}_{\left( {{x_0},{y_0},{z_0}} \right)},{\text{ }}r = \sqrt {17} \cr
& {\text{The standard equation of a sphere of radius }}r,{\text{ centered at}} \cr
& \left( {{x_0},{y_0},{z_0}} \right){\text{ is:}} \cr
& {\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} + {\left( {z - {z_0}} \right)^2} = {r^2} \cr
& {\text{Then,}} \cr
& {\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {z - 2} \right)^2} = {\left( {\sqrt {17} } \right)^2} \cr
& {\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {z - 2} \right)^2} = 17 \cr} $$
