158. Given three mutually perpendicular lines, the distance between any two 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.
158. Introduce a rectangular coordinate system such that the first line coincides with the $x$-axis, the second is parallel to the $y$-axis and passes through the point $(0,0, a)$, and the third is parallel to the $z$-axis and passes through the point $(a, a, 0)$. Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a parallelepiped, where points $A$ and $C$ lie on the first line and have coordinates $\left(x_{1}, 0,0\right),\left(x_{2}, 0,0\right)$, points $B$ and $C_{1}$ lie on the second line, their coordinates are $\left(0, y_{1}, a\right),\left(0, y_{2}, a\right)$, and points $D$ and $B_{1}$ lie on the third line, their coordinates are $\left(a, a, z_{1}\right),\left(a, a, z_{2}\right)$. From the condition of equality of vectors $\vec{A} \vec{D}=\vec{B} \vec{C}=\vec{B}_{1} \vec{C}_{1}$, we get $a-x_{1}=x_{2}=-a$, $a=-y_{1}=y_{2}-a, z_{1}=-a=a-z_{2}$, from which $x_{1}=2 a, x_{2}=-a$, $y_{1}=-a, y_{2}=2 a, z_{1}=-a, z_{2}=2 a$. Thus, $A(2 a, 0,0)$, $B(0,-a, a), C(-a, 0,0), D(a, a,-a), B_{1}(a, a, 2 a), C_{1}(0,2 a, a)$. We can check that $\overrightarrow{A B}=\overrightarrow{D C}$. Further, $|A C|=3 a,|A B|=a \sqrt{6}$, $|B C|=a \sqrt{3}$, i.e., $\triangle A B C$ is a right triangle, so the area of $A B C D$ will be $|A B| \cdot|B C|=3 a^{2} \sqrt{2}$. The equation of the plane $A B C D$ is $y+z=0$, so the distance from $B_{1}$ to it will be $\frac{3 a}{\sqrt{2}}$. Answer: $9 a^{3}$.