7.20. Given three mutually perpendicular lines, the distance between any two of which is a. Find the volume of the parallelepiped, the diagonal of which lies on one line, and the diagonals of two adjacent faces - on the other two lines.
7.20. Consider the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. Let the diagonals of the faces sharing the edge $B C$ lie on given lines, and $A C$ is one of these diagonals. Then $B C_{1}$ is the second such diagonal, and $B_{1} D$ is the diagonal of the parallelepiped lying on the third given line. Introduce a rectangular coordinate system such that the line $A C$ coincides with the $O x$ axis, the line $B C_{1}$ is parallel to the $O y$ axis and passes through the point $(0,0, a)$, and the line $B_{1} D$ is parallel to the $O z$ axis and passes through the point $(a, a, 0)$. Then the points $A$ and $C$ have coordinates $\left(x_{1}, 0,0\right)$ and $\left(x_{2}, 0,0\right)$, the points $B$ and $C_{1}$ have coordinates $\left(0, y_{1}, a\right)$ and $\left(0, y_{2}, a\right)$, and the points $D$ and $B_{1}$ have coordinates $\left(a, a, z_{1}\right)$ and $\left(a, a, z_{2}\right)$. Since $\overrightarrow{A D}=\overrightarrow{B C}=\overrightarrow{B_{1} C_{1}}$, we have $a-x_{1}=x_{2}=-a, a=-y_{1}=y_{2}-a$ and $z_{1}=-a=a-z_{2}$, from which $x_{1}=2 a, x_{2}=-a, y_{1}=-a, y_{2}=-2 a, z_{1}=-a$ and $z_{2}=2 a$. Thus, the coordinates of the vertices $A, B, C, D, B_{1}$ and $C_{1}$ are found.

Simple calculations show that $A C=3 a, A B=a \sqrt{6}$ and $B C=a \sqrt{3}$, i.e., the triangle $A B C$ is a right triangle, and therefore the area of the face $A B C D$ is $A B \cdot B C=3 a^{2} \sqrt{2}$. The plane of the face $A B C D$ is given by the equation $y+z=0$. The distance from the point $\left(x_{0}, y_{0}, z_{0}\right)$ to the plane $p x+q_{y}+r z=0$ is $\left|p x_{0}+q y_{0}+r z_{0}\right| / \sqrt{p^{2}+q^{2}+r^{2}}$ (Problem 1.27), so the distance from the point $B_{1}$ to the face $A B C D$ is $3 a \sqrt{2}$. Therefore, the volume of the parallelepiped is $9 a^{3}$.