7. The height of a regular triangular prism $A B C A_{1} B_{1} C_{1}$ is 6. A sphere $\Omega$ with radius $r=\sqrt{\frac{8}{3}}$ touches all the lateral faces of the prism. Points $M$ and $K$ are chosen on segments $A A_{1}$ and $B B_{1}$, respectively, such that $K M \| A B$, and the planes $A C K$ and $M B_{1} C_{1}$ touch the sphere $\Omega$. Find the volume of the prism and the length of segment $B K$.
Answer: $V=48 \sqrt{3}, B K=5$ or $B K=1$.

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 $2 r \sqrt{3}=4 \sqrt{2}$, the area of the base $S=\frac{\sqrt{3}}{4}(4 \sqrt{2})^{2}=8 \sqrt{3}$, and the volume of the prism is $S h=8 \sqrt{3} \cdot 6=48 \sqrt{3}$.

From the fact that the prism is regular and the sphere touches all its faces, it follows that the plane $A_{1} C_{1} K$ also touches the sphere; the points of tangency of the sphere with the planes $A C K$ and $A_{1} C_{1} K$ lie on the segments $K Q$ and $K Q_{1}$, 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 intersection of the sphere with this plane, be denoted as $O$, and the points of tangency of the segments $K Q_{1}, Q Q_{1}$, and $K Q$ with the circle be denoted as $G, J$, and $P$ respectively. The height $K H$ of the triangle $K Q Q_{1}$ is equal to the height of the triangle lying in the base of the prism, that is, $K H=3 r$.

Write the area $S_{0}$ of the triangle $K Q Q_{1}$ in two ways: $S_{0}=p r$ and $S_{0}=\frac{1}{2} K H \cdot Q Q_{1}=\frac{1}{2} \cdot 3 r \cdot 6=9 r$, where $p$ is the semiperimeter of the triangle $K Q Q_{1}$. Therefore, $p=9$. Let $Q J=x$; then $Q_{1} G=Q_{1} J=6-x$, $P Q=x, K G=K P=\frac{1}{2}(18-2 x-2(6-x))=3$.

Thus, $Q Q_{1}=6, Q K=3+x, Q_{1} K=9-x$. Then Heron's formula gives $S_{0}=\sqrt{9 \cdot 3 \cdot x \cdot(6-x)}$, from which $\sqrt{9 \cdot 3 \cdot x \cdot(6-x)}=9 r, x(6-x)=3 r^{2}$. Substituting the value of the radius from the condition, we get the equation $x^{2}-6 x+8=0$, hence $x=2$ or $x=4$.

Then $Q K=7$ or $Q K=5$, and since $B K=\sqrt{Q K^{2}-B Q^{2}}=\sqrt{Q K^{2}-9 r^{2}}$, we have $B K=5$ or $B K=1$.