7. At the base of the quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ lies a rhombus $A B C D$, in which $B D=12$ and $\angle B A C=60^{\circ}$. A sphere passes through the vertices $D, A, B, B_{1}, C_{1}, D_{1}$.

a) Find the area of the circle obtained by the intersection of the sphere with the plane passing through the points $A_{1}, B_{1}$, and $C_{1}$.

b) Find the angle $A_{1} C B$.

c) Suppose it is additionally known that the radius of the sphere is 8. Find the volume of the prism.
Answer. a) $48 \pi$, b) $90^{\circ}$, c) $192 \sqrt{3}$.

Solution. a) Since the diagonals of a rhombus are the bisectors of its angles, the acute angle of the rhombus is $60^{\circ}$, and $A C=4 \sqrt{3}$. In the cross-section of the sphere by the plane $A_{1} C_{1} B_{1}$, we obtain a circle circumscribed around the triangle $C_{1} B_{1} D_{1}$. The center of this circle is point $A_{1}$, and its radius is equal to the side of the rhombus, which is $4 \sqrt{3}$. Therefore, the area is $48 \pi$.

b) Let $O$ be the center of the sphere. Drop a perpendicular $O H$ from point $O$ to the plane $A B C D$. Then triangles $O H A, O H B$, and $O H D$ are equal by the leg and hypotenuse ( $O H$ is common, $O A=O B=O D$ as radii of the sphere). Therefore, $H A=H B=H D$, so $H$ is the center of the circle circumscribed around triangle $A B D$, i.e., point $H$ coincides with point $C$.

Thus, the segment $O C$ is perpendicular to the plane of the base $A B C D$. Similarly, it can be shown that the segment $O A_{1}$ is perpendicular to the plane $A_{1} B_{1} C_{1} D_{1}$. Therefore, the diagonal $A_{1} C$ is the height of the prism, and the center of the sphere $O$ is its midpoint. Hence, $\angle A_{1} C B=90^{\circ}$.

c) In the right triangle $A O C$, the hypotenuse $A O=8$ and the leg $A C=4 \sqrt{3}$ are known. Therefore, $C O=4$, $A_{1} C=8 ; V=A_{1} C \cdot S_{A B C D}=8 \cdot 24 \sqrt{3}=192 \sqrt{3}$.