10. (20 points) Three solid wooden spheres, with radii of 1, 2, and 3 respectively, are externally tangent to each other. Spheres $O_{1}$, $O_{2}$, and $O_{3}$ are placed between a "V" shaped wooden frame (each sphere is tangent to both sides of the V-shaped frame). Find the degree measure of the angle between the two sides of the V-shaped frame.
10. Let the two faces of the V-shaped wooden frame be $\alpha, \beta$, and the intersection line of $\alpha$ and $\beta$ be $l$, and the plane containing the centers of the three spheres be $\gamma$.
Let $\alpha-l-\beta=\theta$.
Then $l \in \gamma, \alpha-l-\gamma=\beta-l-\gamma=\frac{\theta}{2}$.
Let the projections of points $O_{1}, O_{2}, O_{3}$ on the plane $\alpha$ be $O_{1}^{\prime}, O_{2}^{\prime}, O_{3}^{\prime}$, respectively.
Notice,
$$
\begin{array}{l}
O_{1} O_{2}=3, O_{1} O_{3}=4, O_{2} O_{3}=5, O_{1} O_{1}^{\prime}=1, \\
O_{2} O_{2}^{\prime}=2, O_{3} O_{3}^{\prime}=3,
\end{array}
$$

The plane $O_{1} O_{1}^{\prime} O_{3}^{\prime} O_{3}$ intersects the line $l$ at point $M$.
By $\triangle M O_{1} O_{1}^{\prime} \backsim \triangle M O_{3} O_{3}^{\prime}$, we get
$$
O_{1} M=2, O_{3} M=6 \text {. }
$$

If the plane $O_{2} O_{2}^{\prime} O_{3}^{\prime} O_{3}$ intersects the line $l$ at point $N$, as shown in Figure 5, draw $O_{3} P \perp l$ at point $P$, and connect $O_{3}^{\prime} P$.
Then $O_{3}^{\prime} P \perp l$.
Thus, $\angle O_{3} P O_{3}^{\prime}=\frac{\theta}{2}$.
It is easy to find $O_{3} N=15$.
In $\triangle O_{1} O_{2} O_{3}$, by the cosine rule we get
$$
\begin{array}{l}
\cos \angle O_{1} O_{3} O_{2}=\frac{4}{5} \\
\sin \angle O_{1} O_{3} O_{2}=\frac{3}{5} .
\end{array}
$$

In $\triangle M N O_{3}$, by the cosine rule we get
$$
M N=3 \sqrt{3} \text {. }
$$
$$
\begin{array}{l}
\text { Also, } S_{\triangle O_{3} M N}=\frac{1}{2} M N \cdot O_{3} P \\
=\frac{1}{2} O_{3} M \cdot O_{3} N \sin \angle O_{1} O_{3} O_{2},
\end{array}
$$

Thus, $O_{3} P=\frac{18}{\sqrt{13}}$.
In the right triangle $O_{3} O_{3}^{\prime} P$,
$$
\begin{array}{l}
\sin \frac{\theta}{2}=\sin \angle O_{3} P O_{3}^{\prime}=\frac{O_{3} O_{3}^{\prime}}{O_{3} P}=\frac{\sqrt{13}}{6} \\
\Rightarrow \theta=2 \arcsin \frac{\sqrt{13}}{6}
\end{array}
$$