7. In a right triangular prism, it is known that the base area is $s \mathrm{~m}^{2}$, and the areas of the three lateral faces are $m \mathrm{~m}^{2}, n \mathrm{~m}^{2}, p \mathrm{~m}^{2}$. Then its volume is $\qquad$ $\mathrm{m}^{3}$.
Let the height of a right triangular prism be $h \mathrm{~m}$, and the lengths of the three sides of the base triangle be $a \mathrm{~m}$, $b \mathrm{~m}$, and $c \mathrm{~m}$. Thus,
$$
\begin{array}{l}
a h=m, b h=n, c h=p . \\
\text { Therefore, } S=\frac{1}{4} \sqrt{(a+b+c)(a+b-c)(c+a-b)(b+c-a)} \\
=\frac{1}{4} \sqrt{\frac{m+n+p}{h} \cdot \frac{m+n-p}{h} \cdot \frac{p+m-n}{h} \cdot \frac{n+p-m}{h}} \\
=\frac{1}{4 h^{2}} \sqrt{(m+n+p)(m+n-p)(p+m-n)(n+p-m)} \text {. } \\
\end{array}
$$

Then $h^{2}=\frac{1}{4 S} \sqrt{(m+n+p)(m+n-p)(p+m-n)(n+p-m)}$,
$$
\begin{array}{l}
V^{2}=S^{2} h^{2} \\
=\frac{S}{4} \sqrt{(m+n+p)(m+n-p)(p+m-n)(n+p-m)}, \\
V=\frac{\sqrt{S}}{2} \sqrt[4]{(m+n+p)(m+n-p)(p+m-n)(n+p-m)} .
\end{array}
$$