12.434 The base of a right prism is a parallelogram with an acute angle $\alpha$. The diagonals of the prism form angles $\boldsymbol{\beta}$ and $\boldsymbol{\gamma}(\boldsymbol{\beta}<\boldsymbol{\gamma})$ with the base plane. Find the volume of the prism if its height is $\boldsymbol{H}$.
Solution.

Let the parallelogram $ABCD$ be the base of the given right prism, $\alpha=\angle A$. Then $AC > BD$, $AC = H \cdot \operatorname{ctg} \beta$, $BD = H \cdot \operatorname{ctg} \gamma \quad (AC$ and $BD$ are the projections of the diagonals of the prism). By the cosine theorem for $\triangle ABD$ and $\triangle ABC$, we have:

$$
\left\{\begin{array}{l}
H^{2} \operatorname{ctg}^{2} \gamma = AB^{2} + AD^{2} - 2 \cdot AB \cdot AD \cdot \cos \alpha \\
H^{2} \operatorname{ctg}^{2} \beta = AB^{2} + AD^{2} + 2 \cdot AB \cdot AD \cdot \cos \alpha
\end{array} \Rightarrow\right.
$$

$$
\begin{aligned}
& \Rightarrow AB \cdot AD = \frac{H^{2} \left( \operatorname{ctg}^{2} \beta - \operatorname{ctg}^{2} \gamma \right)}{4 \cos \alpha} \Rightarrow S_{ABCD} = AB \cdot AD \cdot \sin \alpha = \\
& = \frac{H^{2}}{4} \operatorname{tg} \alpha \left( \operatorname{ctg}^{2} \beta - \operatorname{ctg}^{2} \gamma \right) = H^{2} \cdot \frac{\operatorname{tg} \alpha \cdot \sin (\gamma - \beta) \cdot \sin (\gamma + \beta)}{4 \sin^{2} \beta \cdot \sin^{2} \gamma}
\end{aligned}
$$

The volume of the prism is $S_{ABCD} \cdot H$. From this, we get the answer.

Answer: $\quad H^{3} \cdot \frac{\operatorname{tg} \alpha \cdot \sin (\gamma - \beta) \cdot \sin (\gamma + \beta)}{4 \sin^{2} \beta \cdot \sin^{2} \gamma}$.