12.371. In the base of the right prism $A B C A_{1} B_{1} C_{1}\left(A A_{1}\left\|B B_{1}\right\| C C_{1}\right)$ lies an isosceles triangle, where $A B=B C=a$ and $\angle A B C=\alpha$. The height of the prism is $H$. Find the distance from point $A$ to the plane passing through points $B, C$ and $A_{1}$.
Solution.

Draw a perpendicular $AE$ from point $A$ to $BC$ (Fig. 12.231). Then $A_1E \perp BC$ and, consequently, $BC \perp$ plane $AEA_1$. Thus, plane $AEA_1 \perp$ plane $BA_1C$.

In the plane $AEA_1$, drop a perpendicular $AD$ to $A_1E$. Then $AD \perp$ plane $BA_1C$ and the length of the perpendicular $AD$ is the desired distance.

From $\triangle AEB\left(\angle AEB=90^{\circ}\right): AE=AB \sin \angle ABC=a \sin \alpha$.

From $\triangle A_1AE\left(\angle A_1AE=90^{\circ}\right): A_1E=\sqrt{AE^2+AA_1^2}=\sqrt{H^2+a^2 \sin^2 \alpha}$.

$$
S_{\triangle A A_1 E}=\frac{1}{2} A_1 E \cdot A D=\frac{1}{2} A A_1 \cdot A E
$$

Then $A D=\frac{A A_1 \cdot A E}{A_1 E}=\frac{a H \sin \alpha}{\sqrt{H^2+a^2 \sin^2 \alpha}}$.

Answer: $\frac{a H \sin \alpha}{\sqrt{H^2+a^2 \sin^2 \alpha}}$.