108 The shortest distances between a diagonal of a rectangular prism and the edges it does not intersect are $2 \sqrt{5}, \frac{30}{\sqrt{13}}$ and $\frac{15}{\sqrt{10}}$, respectively. Then the volume of this rectangular prism is $\qquad$ .
108 750. As shown in Figure 11, let the common perpendicular line of diagonals $A C_{1}$ and $B B_{1}$ be $E F = d$, and $E F$ lies in the plane perpendicular to $B B_{1}$. This perpendicular plane intersects the cuboid to form a rectangle $E G H I$, and $F$ lies on the intersection line $I G$ of this plane and the plane $A C_{1}$. Since $E F \perp A F$, it follows that $E F \perp$ the projection of $A F$, which is $I F$. Therefore, $d=\frac{a b}{\sqrt{a^{2}+b^{2}}}$, and similarly, other distances can be derived. Thus,
$$
\begin{array}{l}
a b=2 \sqrt{5} \sqrt{a^{2}+b^{2}}, \\
b c=\frac{30}{\sqrt{13}} \sqrt{b^{2}+c^{2}}, \\
a c=\frac{15}{\sqrt{10}} \sqrt{c^{2}+a^{2}} .
\end{array}
$$

Squaring and transforming, we get
$$
\left\{\begin{array}{l}
\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{20}, \\
\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{13}{900}, \\
\frac{1}{c^{2}}+\frac{1}{a^{2}}=\frac{10}{225} .
\end{array}\right.
$$

Solving these equations, we find $a=5, b=10, c=15$, thus
$$
V=a b c=750
$$