12.3.23 $\star \star$ Let $S_{1}$ and $S_{2}$ be two spheres of different radii, externally tangent to each other, and both inside a cone $C$, touching the cone in complete circles. Inside the cone, there are $n$ solid spheres, arranged in a circle, each tangent to the cone $C$, and externally tangent to both $S_{1}$ and $S_{2}$, and each pair of adjacent solid spheres are also externally tangent to each other. Find: all possible values of $n$.
As shown in the figure, let the radii of $S_{1}$ and $S_{2}$ be $R$ and $r$, and the radius of the solid sphere be $x$. Then $D E^{2}=(R+r)^{2}-(R-r)^{2}=4 R r, D E=2 \sqrt{R r}$. Similarly, we get $D F=2 \sqrt{R x}, E F=2 \sqrt{r x}$. From $D E=D F+F E$ we have $\sqrt{R r}=\sqrt{R x}+\sqrt{r x}$, thus $\sqrt{x}=\frac{\sqrt{R r}}{\sqrt{R}+\sqrt{r}}$.
(1) Therefore, the radii of all solid spheres are equal, and their centers are all on a plane perpendicular to the axis of the cone. This is because the projections of the centers on $A B$ are the same point $O$, and these centers are all on a circle with $O$ as the center and $y$ as the radius. Since $S_{\text {quadrilateral DBCE }}=$ $S_{\text {quadrilateral } D B T F}+S_{\text {quadrilateral } E C T E}+S_{\triangle T B C}$, we have $(R+r) \sqrt{R r}=$ $(x+r) \sqrt{x r}+(x+R) \sqrt{R x}+\frac{1}{2}(R+r) y$.
From (1) and (2) we get
$$
\begin{aligned}
\frac{1}{2} y(R+r) & =(R+r) \sqrt{R r}-x \sqrt{x}(\sqrt{R}+\sqrt{r})-r \sqrt{r x}-R \sqrt{R x} \\
& =(R+r) \sqrt{R r}-x \sqrt{R r}-\sqrt{R r}(R+r-\sqrt{R r}) \\
& =R r-x \sqrt{R r}=x(\sqrt{R}+\sqrt{r})^{2}-x \sqrt{R r} \\
& -x(R+r+\sqrt{R r}) .
\end{aligned}
$$

Notice that the centers of the solid spheres form the vertices of a regular $n$-sided polygon inscribed in a circle with $O$ as the center and $y$ as the radius, and the side length of this regular $n$-sided polygon is $2 x$. Thus $\sin \frac{\pi}{n}=\frac{x}{y}=\frac{R+r}{2(R+r+\sqrt{R})}$. Since $\frac{1}{2}>$ $\frac{R+r}{2(R+r+\sqrt{R r})}>\frac{R+r}{3(R+r)}=\frac{1}{3}$. Therefore, $\frac{1}{3}<\sin \frac{\pi}{n}<\frac{1}{2}$. Hence $\frac{\pi}{10}<\arcsin \frac{1}{3}$ $<\frac{\pi}{n}<\arcsin \frac{1}{2}=\frac{\pi}{6}$. Thus $n=7,8,9$.