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 $C D=3$ and $\angle A B D=30^{\circ}$. A sphere passes through the vertices $D, C, B, B_{1}, A_{1}, D_{1}$.

a) Find the area of the circle obtained by the intersection of the sphere with the plane passing through points $A, C$, and $D$.

b) Find the angle $C_{1} A B$.

c) Suppose it is additionally known that the radius of the sphere is 6. Find the volume of the prism.
Answer. a) $9 \pi$, b) $90^{\circ}$, c) 81.

Solution. a) Since the diagonals of a rhombus are the bisectors of its angles, we get that the acute angle of the rhombus is $60^{\circ}$. In the cross-section of the sphere by the plane $A C D$, we obtain a circle circumscribed around the triangle $B C D$. The center of this circle is point $A$, and its radius is equal to the side of the rhombus, which is 3. Therefore, the area is $9 \pi$.

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 $O H C$, $O H B$, and $O H D$ are equal by the leg and hypotenuse ( $O H$ is common, $O C=O B=O D$ as radii of the sphere). Therefore, $H C=H B=H D$, so $H$ is the center of the circle circumscribed around triangle $C B D$, i.e., point $H$ coincides with point $C$.

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

c) In the right triangle $A O C$, the hypotenuse $C O=6$ and the leg $A C=3$ are known. Therefore, $A O=3 \sqrt{3}$, $C_{1} A=6 \sqrt{3} ; V=A C_{1} \cdot S_{A B C D}=6 \sqrt{3} \cdot \frac{9 \sqrt{3}}{2}=81$.