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 the tetrahedron $P-DEF$ is $\qquad$
7. $\frac{\sqrt{17}}{8}$.

Let $P D=x, P E=y, P F=z$.
By the cosine rule, we have
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-x y=7, \\
x^{2}+z^{2}-x z=7, \\
y^{2}+z^{2}-y z=4 .
\end{array}\right.
$$
(1) - (2) gives $y^{2}-z^{2}-x(y-z)=0$.

Since $y \neq z$, we have $y+z=x$.
Substituting into (3) gives
$$
x^{2}-3 y z=4 \text {. }
$$
(1) + (2) gives
$$
\begin{array}{l}
2 x^{2}+y^{2}+z^{2}-x(y+z)=14 \\
\Rightarrow x^{2}+y^{2}+z^{2}=14 \\
\Rightarrow x^{2}+y z=10 .
\end{array}
$$

From (4) and (5), we solve to get $y z=\frac{3}{2}, x=\frac{\sqrt{17}}{\sqrt{2}}$.
Let $\alpha$ be the angle between $P A$ and the plane $P B C$.
It is easy to know that $\sin \alpha=\frac{\sqrt{2}}{\sqrt{3}}$.
$$
\begin{aligned}
\text { Hence } V & =\frac{1}{3} \times \frac{1}{2} y z \sin 60^{\circ} \cdot x \sin \alpha \\
\Rightarrow V & =\frac{1}{3} \times \frac{1}{2} \times \frac{3}{2} \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{17}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{3}} \\
& =\frac{\sqrt{17}}{8} .
\end{aligned}
$$