5. Given a regular tetrahedron $ABCD$ with edge length $a(a>3)$ as shown in the figure, points $E, F, G$ are on edges $AB, AC, AD$ respectively. Then the number of $\triangle EFG$ that satisfies $EF=EG=3, FG=2$ is
A. 1
B. 2
C. 3
D. 4
Let $A E=x, A F=y, A G=z$. According to the problem, we have $\left\{\begin{array}{l}x^{2}+y^{2}-x y=9, \\ x^{2}+z^{2}-x z=9, \\ y^{2}+z^{2}-y z=4,\end{array}\right.$
(1) - (2) $\Rightarrow(y-z)(y+z-x)=0 \Rightarrow y=z$ or $y+z=x$.

If $y=z$, we can solve to get $x=\sqrt{6}+1, y=z=2$; if $x=y+z$ we can get $\left\{\begin{array}{l}y^{2}+z^{2}=\frac{13}{2}, \\ y z=\frac{5}{2} .\end{array}\right.$ In summary, $\triangle E F G$ has 3 solutions, so the answer is $C$.