Chapter 9 - Section 9.3 - Three-Dimensional Coordinate Geometry - 9.3 Exercises - Page 652: 16

center $(-2, 3, -1)$, radius $2\sqrt{6}$

$x^2+y^2+z^2+4x-6y+2z=10$ $x^2+4x+y^2-6y+z^2+2z=10$ $x^2+4x+(\frac{4}{2})^2+y^2-6y+(\frac{-6}{2})^2+z^2+2z+(\frac{2}{2})^2=10+(\frac{4}{2})^2+(\frac{-6}{2})^2+(\frac{2}{2})^2$ $x^2+4x+4+y^2-6y+9+z^2+2z+1=10+4+9+1$ $(x+2)^2+(y-3)^2+(z+1)^2=24$ This is a sphere with center $(-2, 3, -1)$ and radius $\sqrt{24}=2\sqrt{6}$.