7. Given a right triangular prism $A B C A_{1} B_{1} C_{1}$. A sphere with diameter $A C$ intersects the edges $A B$ and $B C$ at points $F$ and $N$, respectively, different from the vertices of the prism. Segments $C_{1} F$ and $A_{1} N$ intersect at point $P$, and at the same time $A_{1} N=7, C_{1} P=6$.

a) Find the angle $P F A$.

b) Find the ratio $A F: F B$.

c) Suppose it is additionally known that $A B=6$. Find the volume of the prism.
Answer: a) $90^{\circ}$, b) $5: 1$, c) $V=21 \sqrt{10}$.

Solution. a) Points $F$ and $N$ lie on the circle with diameter $A C$; therefore, $\angle A F C=90^{\circ}, \angle A N C=90^{\circ}$ (i.e., $A N$ and $C F$ are altitudes of triangle $A B C$). The line $C F$ is the projection of the line $C_{1} F$ onto the base plane, and since $C F \perp A B$, by the theorem of three perpendiculars, $C_{1} F \perp A B$, i.e., $\angle P F A=90^{\circ}$.

b) Since the lines $C_{1} F$ and $A_{1} N$ intersect, all four points $A_{1}, C_{1}, F$, and $N$ lie in the same plane (let's call it $\alpha$). Therefore, the lines $F N$ and $A_{1} C_{1}$ lie in the same plane $\alpha$, and since they do not intersect (as they lie in the parallel bases of the prism), $F N \| A_{1} C_{1}$. Thus, $F N \| A C$. The trapezoid $A F N C$ is inscribed in a circle, so it is isosceles, and the angles at its base $A C$ are equal, which means triangle $A B C$ is isosceles $(A B=B C)$.

Triangles $A_{1} C_{1} P$ and $F N P$ are similar by two angles. From the equality of triangles $C C_{1} F$ and $A A_{1} N$, it follows that $A_{1} N=C_{1} F$, so both triangles $A_{1} C_{1} P$ and $F N P$ are isosceles with bases $A_{1} C_{1}$ and $F N$ respectively. Therefore, $F N: A C=F N: A_{1} C_{1}=F P: P C_{1}=\left(F C_{1}-P C_{1}\right): P C_{1}=\left(N A_{1}-P C_{1}\right): P C_{1}=1: 6$, hence $\frac{A F}{B F}=\frac{A B-B F}{B F}=\frac{A B}{B F}-1=\frac{A C}{F N}-1=6-1=5$.

c) If $A B=6$, then $B F=1, A F=5, B C=6 ; C F=\sqrt{B C^{2}-B F^{2}}=\sqrt{35} ; C C_{1}=\sqrt{C_{1} F^{2}-C F^{2}}=\sqrt{14}$. Therefore, the area of the base of the prism is $S_{A B C}=\frac{1}{2} \cdot 6 \cdot \sqrt{14}$, and the volume of the prism is $V=S_{A B C} \cdot C C_{1}=21 \sqrt{10}$.