Solution: Let $(x, y, z)$ be the center of the sphere. By the given condition, we have
$$(x-2)^{2} + y^{2} + z^{2} = x^{2} + (y-4)^{2} + z^{2} = x^{2} + y^{2} + (z-6)^{2}.$$
Subtracting $x^{2} + y^{2} + z^{2}$ yields
$$x^{2} - (x-2)^{2} = y^{2} - (y-4)^{2} = z^{2} - (z-6)^{2},$$
or
$$4(x-1) = 8(y-2) = 12(z-3)$$
Therefore $(x-1, y-2, z-3)$ must be $(6t, 3t, 2t)$ for some integer $t$. Checking small values of $t$ yields that the possibilities for $(x, y, z)$ closest to $(2,0,0)$ are $(-5,-1,1)$, $(1,2,3)$, and $(7,5,5)$. The second yields a sphere that passes through the origin and is thus forbidden. The other two yield $r^{2} = 51$ and $r^{2} = 75$, so 51 is the answer.