smc

The two equations of the balls are $$x^2+y^2+{\left(z-\frac{21}{2}\right)}^2\le 36$$ $$x^2+y^2+(z-1{)}^2\le \frac{81}{4}$$ Note that along the $z$ axis, the first ball goes from $10.5\pm 6$, and the second ball goes from $1\pm 4.5$. The only integer value that $z$ can be is $z=5$. Plugging that in to both equations, we get: $$x^2+y^2\le \frac{23}{4}$$ $$x^2+y^2\le \frac{17}{4}$$ The second inequality implies the first inequality, so the only condition that matters is the second inequality. From here, we do casework, noting that $|x|,|y|\le 3$: For $x=\pm 2$, we must have $y=0$. This gives $2$ points. For $x=\pm 1$, we can have $y\in \{-1,0,1\}$. This gives $2\cdot 3=6$ points. For $x=0$, we can have $y\in \{-2,-1,0,1,2\}$. This gives $5$ points. Thus, there are $\boxed{13}$ possible points, giving answer $\boxed{13}$.