7. The height of a regular triangular prism $A B C A_{1} B_{1} C_{1}$ is 12. A sphere $\Omega$ with radius $r=\sqrt{\frac{35}{3}}$ touches all the lateral faces of the prism. Points $K$ and $L$ are chosen on segments $A A_{1}$ and $B B_{1}$, respectively, such that $K L \| A B$, and the planes $K B C$ and $L A_{1} C_{1}$ touch the sphere $\Omega$. Find the volume of the prism and the length of segment $A K$.
Answer: $V=420 \sqrt{3}, A K=8$ or $A K=4$.

Solution. Since the sphere $\Omega$ of radius $r$ touches all the lateral faces of the prism, circles of the same radius $r$ can be inscribed in the bases of the prism. Therefore, the side of the base is $2 r \sqrt{3}=2 \sqrt{35}$, the area of the base $S=\frac{\sqrt{3}}{4}(2 \sqrt{35})^{2}=35 \sqrt{3}$, and the volume of the prism is $S h=35 \sqrt{3} \cdot 12=420 \sqrt{3}$.

From the fact that the prism is regular and the sphere touches all its faces, it follows that the plane $L A C$ also touches the sphere; in this case, the points of tangency of the sphere with the planes $L A C$ and $L A_{1} C_{1}$ lie on the segments $L Q$ and $L Q_{1}$, where points $Q$ and $Q_{1}$ are the midpoints of the edges $A C$ and $A_{1} C_{1}$.

Consider the plane of the rectangle $B B_{1} Q_{1} Q$. Let the center of the circle $\omega$, obtained by the section of the sphere by this plane, be denoted by $O$, and the points of tangency of the segments $L Q_{1}, L Q$, and $Q Q_{1}$ with the circle be denoted by $G, P$, and $J$ respectively. The height $L H$ of the triangle $L Q Q_{1}$ is equal to the height of the triangle lying in the base of the prism, that is, $L H=3 r$.

We can write the area $S_{0}$ of the triangle $L Q Q_{1}$ in two ways: $S_{0}=p r$ and $S_{0}=\frac{1}{2} L H \cdot Q Q_{1}=\frac{1}{2} \cdot 3 r \cdot 12=18 r$, where $p$ is the semiperimeter of the triangle $L Q Q_{1}$. Therefore, $p=18$. Let $Q J=x$; then $Q_{1} G=Q_{1} J=12-x$, $P Q=x, L G=L P=\frac{1}{2}(36-2 x-2(12-x))=6$.

Thus, $Q Q_{1}=12, Q L=6+x, Q_{1} L=18-x$. Then Heron's formula gives $S_{0}=\sqrt{18 \cdot 6 \cdot x \cdot(12-x)}$, from which $\sqrt{18 \cdot 6 \cdot x \cdot(12-x)}=18 r, x(12-x)=3 r^{2}$. Substituting the value of the radius from the condition, we get the equation $x^{2}-12 x+35=0$, hence $x=5$ or $x=7$.

Then $Q L=13$ or $Q L=11$, and since $B L=\sqrt{Q L^{2}-B Q^{2}}=\sqrt{Q L^{2}-9 r^{2}}$, we have $B L=4$ or $B L=8$. It remains to note that $A K=B L=Q H$.