7. In a regular tetrahedron $P-ABC$, points $D, E, F$ are on edges $PA, PB, PC$ respectively. If $PE \neq PF$, and $DE = DF = \sqrt{7}, EF = 2$, then the volume of tetrahedron $P-DEF$ is $\qquad$.
Solve:
As shown in the figure, let $P D=x, P E=y, P F=z$, then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-x y=7 \\
x^{2}+z^{2}-x z=7, \\
y^{2}+z^{2}-y z=4,
\end{array}\right.
$$
(1) - (2) gives $y^{2}-z^{2}-x(y-z)=0 \Rightarrow x=y+z$.

Thus $\left\{\begin{array}{l}y^{2}+z^{2}+y z=7, \\ y^{2}+z^{2}-y z=4,\end{array} \Rightarrow\left\{\begin{array}{l}y^{2}+z^{2}=\frac{11}{2}, \\ y z=\frac{3}{2},\end{array}\right.\right.$ $\Rightarrow x=\frac{\sqrt{34}}{2}, y=\frac{\sqrt{34}+\sqrt{10}}{4}, z=\frac{\sqrt{34}-\sqrt{10}}{4}$. Since the volume of a regular tetrahedron $V=\frac{\sqrt{2}}{12} a^{3}$, we have $\frac{V_{P-D E F}}{V_{P-A B C}}=\frac{x y z}{a^{3}} \Rightarrow V_{P-D E F}=\frac{\sqrt{2}}{12} x y z=\frac{\sqrt{2}}{12} \cdot \frac{\sqrt{34}}{2} \cdot \frac{3}{2}=\frac{\sqrt{17}}{8}$.