5. In a regular tetrahedron $ABCD$, $E$ and $F$ are on edges $AB$ and $AC$ respectively, satisfying $BE=3$, $EF=4$, and $EF$ is parallel to plane $BCD$. Then the area of $\triangle DEF$ is $\qquad$.
Answer: $2 \sqrt{33}$.
Solution: From the given conditions, we know that $E F$ is parallel to $B C$. Since the faces of the regular tetrahedron $A B C D$ are congruent equilateral triangles, we have
$$
A E=A F=E F=4, A D=A B=A E+B E=7 .
$$

By the cosine rule,
$$
\begin{aligned}
D E & =\sqrt{A D^{2}+A E^{2}-2 A D \cdot A E \cdot \cos 60^{\circ}} \\
& =\sqrt{49+16-28}=\sqrt{37},
\end{aligned}
$$

Similarly, $D F=\sqrt{37}$.
Construct the altitude $D H$ from the vertex $D$ to the base $E F$ of the isosceles $\triangle D E F$, then $E H=\frac{1}{2} E F=2$, hence
$$
D H=\sqrt{D E^{2}-E H^{2}}=\sqrt{33},
$$

Therefore, $S_{\triangle D E F}=\frac{1}{2} \cdot E F \cdot D H=2 \sqrt{33}$.