Let the radii of $S_1$ and $S_2$ be $a>b$. Then by considering a plane through the axis of the cone that intersects one of the solid spheres, say $S_3$, in a maximal circle, we get by Descartes' theorem for the radius $r$ of $S_3$ $\frac1{\sqrt{r}}=\frac1{\sqrt{a}}+\frac1{\sqrt{b}}$ or $r=\frac{ab}{(\sqrt{a}+\sqrt{b})^2}$. For the (half) opening angle $\phi$ of the cone, we have $\sin\phi=\frac{a-b}{a+b}$. Define $\psi$ by $\sin\psi=\frac{b-r}{b+r}$ (this is the corresponding angle intervening between $S_2$ and $S_3$). Let $R$ denote the distance of the center of $S_3$ from the axis of the cone. Then we have \[R=(b+r)\sin(\phi+\psi)=2\frac{(a-b)\sqrt{br}+(b-r)\sqrt{ab}}{a+b}=\frac{2ab(a+b+\sqrt{ab})}{(a+b)(\sqrt{a}+\sqrt{b})^2}\] So $\frac rR=\frac{a+b}{2(a+b+\sqrt{ab})}=\frac{1}{2+\frac{2\sqrt{x}}{1+x}}$ where $x=\frac ba$. This function is decreasing in $(0,1)$ with values between $\frac12$ and $\frac13$. As we want $\frac rR=\sin\frac\pi n$, we get $n=7,8,9$. The limit case $n=6$ (for $x\to 0$) cannot be attained.