Answer
$(x-3)^{2}$ $+$ $(y-5)^{2}$ $+$ $(z+3)^{2}$ $=$ $41$
Work Step by Step
Because the diameters has endpoints $($ $5$, $4$, $3$ $)$ and $($ $1$, $6$, $-$$9$ $)$,
so the center of the sphere is
$($ $\frac{5+1}{2}$, $\frac{4+6}{2}$, $\frac{3+(-9)}{2}$$)$ $=$ $($ $3$, $5$, $-3$ $)$
the radius of the sphere is
$r$ $=$ $\sqrt{(5-3)^{2}+(4-5)^{2}+(3-(-3))^{2}}$ $=$ $\sqrt {41}$
Therefore, the equation of the sphere is
$(x-3)^{2}$ $+$ $(y-5)^{2}$ $+$ $(z+3)^{2}$ $=$ $41$