5. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, the angle between $A B_{1}$ and $A_{1} D$ is $\alpha$, the angle between $A C$ and $B C_{1}$ is $\beta$, and the angle between $A_{1} C_{1}$ and $C D_{1}$ is $\gamma$. Therefore, $\alpha+\beta+\gamma=$ $\qquad$ .
5. As shown in the figure, connect $A_{1} D, A D_{1}, D C_{1}, C D_{1}, C_{1} B, A B_{1}$.

Since $D_{1} D \perp A D, D_{1} D \perp D C, A D \perp D C$, by using the cosine rule, it is easy to find that the three interior angles of $\triangle A D_{1} C$ are all acute angles. Similarly, it can be proven that the three interior angles of $\triangle A_{1} D C$ are also all acute angles.

Because $A B_{1} / / D C_{1}$, $\angle A_{1} D C_{1}$ is the angle formed by $A B_{1}$ and $A_{1} D$, i.e., $\angle A_{1} D C_{1}=\alpha$. Similarly, $\angle D_{1} A C=\beta, \angle A C D_{1}=\gamma$. Clearly, $\triangle A D_{1} C \cong \triangle A_{1} D C_{1}$, so $\angle A D_{1} C=\angle A D C_{1}=\alpha$. In $\triangle A D_{1} C$, $\angle A D_{1} C = \angle D_{1} A C + \angle D_{1} C A = 180^{\circ}$,
thus $\alpha+\beta+\gamma=180^{\circ}$.