Chapter 12 - Vectors and the Geometry of Space - 12.1 Three-Dimensional Coordinate Systems - 12.1 Exercises - Page 837: 16

$(x-1)^2+(y-2)^2+(z-3)^2=14$

Because it is a sphere, the equation is of the form $(x-a)^2+(y-b)^2+(z-c)^2=r^2$. Because the center is $ (1,2,3)$, it is of the form $(x-1)^2+(y-2)^2+(z-3)^2=r^2$. Because it passes through the origin, we can plug in $(0,0,0)$ and find that $r^2=14$