Answer
center $(0, 7, -3)$, radius $\sqrt{58}$
Work Step by Step
$x^2+y^2+z^2=14y-6z$
$x^2+y^2-14y+z^2+6z=0$
$x^2+y^2-14y+(\frac{-14}{2})^2+z^2+6z+(\frac{6}{2})^2=0+(\frac{-14}{2})^2+(\frac{6}{2})^2$
$x^2+y^2-14y+49+z^2+6z+9=49+9$
$x^2+(y-7)^2+(z+3)^2=58$
This is a sphere with center $(0, 7, -3)$ and radius $\sqrt{58}$.
