255. Two trihedral angles are arranged in such a way that the vertex of one is equidistant from the faces of the other and vice versa, the distance between the vertices is \(a\). What is the minimum volume of the hexahedron bounded by the faces of these angles, if all the plane angles of one of them are \(60^{\circ}\), and the other - \(90^{\circ}\)?
255. Let $A$ be the vertex of a trihedral angle, the plane angles of which are right angles, and $B$ be the vertex of another angle. Take a point $M$ on the segment $AB$ such that $2|AM| = |MB|$. Draw a plane through point $M$ perpendicular to $AB$. The drawn plane will intersect each of these two trihedral angles in an equilateral triangle with side $b = a \sqrt{\frac{2}{3}}$. In Fig. 52, $\triangle PQR$ corresponds to the section of the trihedral angle with vertex $A$. The face $BCD$ cuts off a pyramid $QFKL$ from the pyramid $APQR$ (where the point $F$ is clear from Fig. 52, b). The volume of this pyramid is proportional to the product $|QK| \cdot |QL| \cdot |QF|$. The value of $|QF|$ obviously reaches its maximum when $\alpha = \pi / 3$, where $\alpha = \widehat{CMQ}$. We will prove that $|KQ| \times |QL|$ also reaches its maximum value when $\alpha = \pi / 3$. Since $KL$ is tangent to the inscribed circle in $\triangle PQR$, the perimeter of $\triangle KQL$ is constant and equal to $b$. Let $|KQ| = x$ and $|QL| = y$, then $|KL| = b - x - y$. Write the cosine theorem for $\triangle KQL$:

$$
(b - x - y)^2 = x^2 + y^2 - xy \Rightarrow b^2 - 2b(x + y) + 3xy = 0 \Rightarrow
$$

$$
\Rightarrow b^2 - 4b \sqrt{xy} + 3xy \geq 0
$$

Thus, either $\sqrt{xy} \leq \frac{b}{3}$ or $\sqrt{xy} \geq b$. But $0 \leq x \leq \frac{b}{2}$ and $0 \leq y \leq \frac{b}{2}$. Therefore, $\sqrt{xy} \leq \frac{b}{3}$. Equality is achieved if $x = y = \frac{b}{3}$.

Thus, the volume of the pyramid $QKLF$ is maximum when $\alpha = \pi / 3$. In this case, $|KQ| = |QL| = \frac{b}{3} = \frac{a}{3} \sqrt{\frac{2}{3}}$. Further, when $\alpha = \pi / 3$, $N$ is the midpoint of $QM$ (Fig. 52, b). Drawing $QT$ parallel to $FB$, we get $|BT| = |MB|$. Thus,

$$
\frac{|AF|}{|FQ|} = \frac{|AB|}{|BT|} = \frac{3}{2}, \quad |QF| = \frac{2}{5}|AQ|
$$

The volume of the pyramid $APQR$ is easily found, it is $\frac{a^3 \sqrt{3}}{54}$. From the pyramid $APQR$, three pyramids equal to the pyramid $QFKL$ are cut off.

The volume of each of them is $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{5} = \frac{2}{45}$ of the volume of $APQR$. Thus, when $\alpha = \pi / 3$, the remaining polyhedron from the pyramid $APQR$ has a volume of

$$
\frac{a^3 \sqrt{3}}{54} \left(1 - \frac{2}{15}\right) = \frac{13 a^3 \sqrt{3}}{810}
$$

Reasoning in the same way, we get that when $\alpha = \pi / 3$, the remaining polyhedron from the pyramid $BCDE$ has the smallest volume, and the volume of this polyhedron will be $\frac{11 a^3 \sqrt{3}}{324}$.

Adding the obtained volumes, we get the answer $\frac{a^3 \sqrt{3}}{20}$. 256. Let $|BD| = 2x$. It is not difficult to find

$$
V = V_{ABCD} = \frac{x |1 - 2x^2| \sqrt{3 - 4x^2}}{6(1 - x^2)}
$$

Substituting $u = 1 - x^2$ and then $w = 4u + 1/u$, we get

$$
\begin{aligned}
(6V)^2 = \frac{x^2 (1 - 2x^2)^2 (3 - 4x^2)}{(1 - x^2)^2} & = \\
= \frac{(1 - u)(2u - 1)^2 (4u - 1)}{u^2} & = \left(5 - \frac{1}{u} - 4u\right) \left(4u + \frac{1}{u} - 4\right) = \\
& = (5 - w)(w - 4) = -w^2 + 9w - 20
\end{aligned}
$$

The maximum value is achieved when $w = 9/2$, from which

$$
x = \sqrt{1 - u} = \sqrt{1 - \frac{9 \pm \sqrt{17}}{16}}
$$

Answer: the maximum value of $V_{ABCD}$ is $\frac{1}{12}$.