3. Let $\alpha$, $\beta$, and $\gamma$ be the angles formed by the diagonal of a rectangular prism with the three faces sharing a common vertex. Then the range of $\alpha+\beta+\gamma$ is $\qquad$ .
3. $\left(\frac{\pi}{2}, 3 \arcsin \frac{\sqrt{3}}{3}\right]$.

In a rectangular parallelepiped, there is
$$
\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1 \text {. }
$$

Notice that
$$
\begin{array}{l}
\sin ^{2} \alpha=1-\sin ^{2} \beta-\sin ^{2} \gamma=\cos ^{2} \beta-\sin ^{2} \gamma \\
=\frac{1}{2}(1+\cos 2 \beta)-\frac{1}{2}(1-\cos 2 \gamma) \\
=\cos (\beta+\gamma) \cdot \cos (\beta-\gamma)>0 .
\end{array}
$$

Since $\beta, \gamma$ are both acute angles, we have
$$
\cos (\beta-\gamma)>0 \text {. }
$$

Thus, $\cos (\beta+\gamma)>0$, which means $\beta+\gamma<\frac{\pi}{2}-\beta-\gamma \text {. }$

Therefore, $\alpha+\beta+\gamma>\frac{\pi}{2}$.
When $\alpha \rightarrow 0^{+}, \beta \rightarrow 0^{+}, \gamma \rightarrow \frac{\pi}{2}$, we have
$$
\alpha+\beta+\gamma \rightarrow \frac{\pi}{2} \text {. }
$$

In summary, $\frac{\pi}{2}<\alpha+\beta+\gamma \leqslant 3 \arcsin \frac{\sqrt{3}}{3}$.