7. In a regular tetrahedron $ABCD$, points $E$ and $F$ are on edges $AB$ and $AC$, respectively, satisfying $BE=3$, $EF=4$, and $EF \parallel$ plane $BCD$. Then the area of $\triangle DEF$ is $\qquad$.
7. $2 \sqrt{33}$.
From the condition, we know $E F / / B C$.
Since each face of the regular tetrahedron $A B C D$ is an equilateral triangle, we have,
$$
\begin{array}{l}
A E=A F=E F=4, \\
A D=A B=A E+B E=7 .
\end{array}
$$

By the cosine rule,
$$
D E=\sqrt{A D^{2}+A E^{2}-2 A D \cdot A E \cos 60^{\circ}}=\sqrt{37} \text {. }
$$

Similarly, $D F=\sqrt{37}$.
As shown in Figure 1, draw the altitude $D H$ from the vertex $D$ to the base $E F$ of the isosceles $\triangle D E F$.
$$
\begin{array}{l}
\text { Then } E H=\frac{1}{2} E F=2 \\
\Rightarrow D H=\sqrt{D E^{2}-E H^{2}}=\sqrt{33} \\
\Rightarrow S_{\triangle D E F}=\frac{1}{2} E F \cdot D H=2 \sqrt{33} .
\end{array}
$$