84. In a regular prism $A B C A_{1} B_{1} C_{1}$, the length of the lateral edge and the height of the base are equal to $a$. Through vertex $A$, two planes are drawn: one perpendicular to the line $A B_{1}$, and the second perpendicular to the line $A C_{1}$. Through vertex $A_{1}$, two more planes are drawn: one perpendicular to the line $A_{1} B$, and the second perpendicular to the line $A_{1} C$. Find the volume of the polyhedron bounded by these four planes and the plane $B B_{1} C_{1} C$.
84. Let's draw a plane through the edge $A A_{\mathrm{i}}$ perpendicular to the plane $B C C_{1} B_{1}$ (Fig. 6). $M$ and $N$ are the points of intersection of this plane with $C_{1} B_{1}$ and $C B_{*}$. Take a point $K$ on $M N$ such that $|N K|=|M N|$. According to the condition, $A A_{1}{ }^{*} M N$ is a square, so $A K$ is perpendicular to $A M$,

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-059.jpg?height=518&width=413&top_left_y=644&top_left_x=153)

Fig. 6.

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-059.jpg?height=447&width=345&top_left_y=716&top_left_x=761)

Fig. 7.

From this, it follows that $A K$ is perpendicular to the plane $A C_{1} B_{1}$, i.e., $A K$ is the line of intersection of the planes passing through vertex $A$. Similarly, for vertex $A_{1}$, we determine the point $L$. The lines $A K$ and $A_{1} L$ intersect at point $S$. Thus, our polyhedron is a quadrilateral pyramid $S K P L Q$ with vertex $S$, and its base lies in the plane $B B_{1} C_{1} C$. Further, $B_{1} N$ is the projection of $A B_{1}$. Therefore, the plane passing through $A$ and perpendicular to $A B_{1}$ intersects the plane $B B_{1} C_{1} C$ along a line perpendicular to $B_{1} N$. From the condition, it follows that $\triangle B_{1} N C_{1}$ is equilateral. Thus, the quadrilateral $P L Q K$, which is the base of the pyramid $S P L Q K$, is a rhombus composed of two equilateral triangles with side length $|K L|=3 a$.

$$
\text { Answer: } \frac{9 a^{3} \sqrt{3}}{4} \text {. }
$$