7. Given a regular prism $A B C D A_{1} B_{1} C_{1} D_{1}$ with base $A B C D$. Planes $\alpha$ and $\beta$ are perpendicular to $B_{1} D$ and pass through vertices $A$ and $D_{1}$, respectively. Let $F$ and $H$ be the points of intersection of planes $\alpha$ and $\beta$ with the diagonal $B_{1} D$, such that $D F < D H$.

a) Find the ratio $B_{1} H: D F$.

b) Suppose it is additionally known that a sphere of radius 3 touches all the lateral faces of the prism, as well as planes $\alpha$ and $\beta$. Find the segment $B_{1} D$ and the volume of the prism $A B C D A_{1} B_{1} C_{1} D_{1}$.
Answer: a) $2: 1$, b) $B_{1} D=3(1+\sqrt{13}), V=108 \sqrt{6+2 \sqrt{13}}$.

Solution. a) By symmetry (with respect to the plane $B D D_{1} B_{1}$), the plane $\alpha$ passes through the point $C$ - and, therefore, through the center $O$ of the face $A B C D$. Segments $B_{1} H$ and $D F$ are projections of parallel segments $B_{1} D_{1}$ and $D O$ onto the line $B_{1} D$, and $B_{1} D_{1}=2 D O$. Therefore, $B_{1} H: D F=2$.

b) Since the sphere is tangent to all lateral faces of the prism, its projection onto the base is a circle inscribed in this base. Thus, $A B=2 r=6$. Moreover, $\alpha$ and $\beta$ are two parallel planes tangent to the sphere, so the distance between them is equal to the diameter of the sphere, which is 6. Since $B_{1} D \perp \alpha$, this distance is the segment $H F$, so $H F=6$.

Let $B_{1} D=d$. Since $D_{1} H$ is the height of the right triangle $B_{1} D_{1} D$, then $B_{1} H \cdot B_{1} D=B_{1} D_{1}^{2}=72$ and, consequently, $B_{1} H=\frac{72}{d}$. Then $D F=\frac{1}{2} B_{1} H=\frac{36}{d}$ and $H F=B_{1} D-B_{1} H-D F=d-\frac{72}{d}-\frac{36}{d}$. We get the equation $6=d-\frac{108}{d}$, from which $d^{2}-6 d-108=0, d=3+3 \sqrt{13}$ (since $d>0$ ).

Finally, the height of the prism is $h=\sqrt{B_{1} D^{2}-B D^{2}}=\sqrt{9(14+2 \sqrt{13})-72}=3 \sqrt{6+2 \sqrt{13}}$; then the volume of the prism is $V=A B^{2} \cdot h=108 \sqrt{6+2 \sqrt{13}}$.