Answer
$(x)^2 + (y+2)^2 + (z-6)^2 \leq 41$
Work Step by Step
Equation of a sphere with center $(a,b,c)$ and radius $r$.
$(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$
Plugging in the point $(0,-2,6)$
$(x-0)^2 + (y+2)^2 + (z-6)^2 = r^2$
Plugging in the point $(1,4,8)$
$(1)^2 + (4+2)^2 + (8-6)^2 = 1 + 36 + 4 = 41$
Equation of Sphere:
$(x)^2 + (y+2)^2 + (z-6)^2 = 41$
Since it's a ball (not a hollow center):
$(x)^2 + (y+2)^2 + (z-6)^2 \leq 41$
