10. The edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is 3, and points $P, Q, R$ are on the edges $A A_{1}, B B_{1}, C_{1} D_{1}$ respectively, with $A_{1} P=1, B_{1} Q=2, C_{1} R=1$. Let the dihedral angle (acute) between plane $P Q R$ and plane $A B C D$ be $\alpha$, find $\alpha$.
10. Draw a perpendicular from $R$ to the plane $ABCD$, then the foot of the perpendicular $R^{\prime}$ lies on the line segment $CD$, and $C R^{\prime}=1$.

Since $\triangle A R^{\prime} B$ is the projection of $\triangle P R Q$ on the plane $ABCD$, and it can be found that $S_{\triangle A B R}=\frac{1}{2} A B^{2}=\frac{9}{2}, P Q=\sqrt{10}, P R=$ $\sqrt{14}, Q R=\sqrt{14}$, the height from the vertex to the base of the isosceles $\triangle P Q R$ is $\frac{\sqrt{46}}{2}$, and $S_{\triangle P Q R}=\frac{1}{2} \sqrt{115}$. Therefore, $\cos \alpha=\frac{9}{2} / \frac{\sqrt{115}}{2}=$ $\frac{9}{115} \sqrt{115}$. Hence, $\alpha=\arccos \frac{9 \sqrt{115}}{115}$ is the required angle.