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 $(6 t, 3 t, 2 t)$ 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.