10,11

Three spheres with radii $\sqrt{6}$, 1, and 1 touch each other pairwise. A plane $\gamma$ is drawn through the line containing the centers $A$ and $B$ of the second and third spheres, such that the center $O$ of the first sphere is at a distance of 1 from this plane. Find the angle between the projections of the lines $OA$ and $OB$ onto the plane $\gamma$ and compare it with $\arccos \frac{4}{5}$.
Since the second and third spheres are equal, they cannot touch internally. Therefore, $A B=2$. If the first sphere touches one of the equal spheres externally and the second one internally, then the centers of the three spheres lie on the same straight line. Then the plane $\gamma$ passes through the center $O$ of the first sphere, which contradicts the problem's condition.

Let the spheres with centers $A$ and $B$ touch the sphere with center $O$ externally. Then $O A=O B=1+\sqrt{6}$.

Drop a perpendicular $O H$ from point $O$ to the plane $\gamma$. From the right triangles $O A H$ and $O B H$, we find that

$$
B H=A H=\sqrt{O A^{2}-O H^{2}}=\sqrt{(1+\sqrt{6})^{2}-1}=\sqrt{6+2 \sqrt{6}}
$$

Then, by the cosine rule,

$$
\begin{gathered}
\cos \angle A H B=\frac{A H^{2}+B H^{2}-A B^{2}}{2 A H \cdot B H}=\frac{2(6+2 \sqrt{6})-4}{2(6+2 \sqrt{6})}= \\
=\frac{2+\sqrt{6}}{3+\sqrt{6}}=\frac{\sqrt{2}(\sqrt{2}+\sqrt{3})}{\sqrt{3}(\sqrt{3}+\sqrt{2})}=\sqrt{\frac{2}{3}}
\end{gathered}
$$

Let the spheres with centers $A$ and $B$ touch the sphere with center $O$ internally. Then $O A=O B=\sqrt{6}-1$ and

$$
\sqrt{\frac{2}{3}}=\sqrt{\frac{16}{24}} \sqrt{\frac{16}{25}}=\frac{4}{5}
$$

then $\arccos \sqrt{\frac{2}{3}}<\arccos \frac{4}{5}$.

## Answer

$\arccos \sqrt{\frac{2}{3}}$