3. Given the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ with its right focus at $F$, and $P$ is a point on the ellipse, point $A(0,2 \sqrt{3})$. When the perimeter of $\triangle A P F$ is maximized, the area of $\triangle A P F$ is $\qquad$.
3. $\frac{21 \sqrt{3}}{4}$.

Notice that, the perimeter of $\triangle A P F$ is
$$
\begin{array}{l}
|A F|+|A P|+|P F| \\
=|A F|+2 a-\left|P F^{\prime}\right|+|A P| .
\end{array}
$$

To maximize the perimeter, we need to maximize $|A P|-\left|P F^{\prime}\right|$. As shown in Figure 2, this is achieved when $A, P, F^{\prime}$ are collinear. Given point $A(0,2 \sqrt{3})$, we know $c=2$.
Then the slope angle of line $A P$ is $\frac{\pi}{3}$,
$$
\angle P F^{\prime} F=\frac{2 \pi}{3} \text {. }
$$

By the cosine rule,
$$
\begin{array}{l}
\left(6-\left|P F^{\prime}\right|\right)^{2}=\left|P F^{\prime}\right|^{2}+16+4\left|P F^{\prime}\right| \\
\Rightarrow\left|P F^{\prime}\right|=\frac{5}{4} \\
\Rightarrow S_{\triangle A P F}=\frac{1}{2}|P A|\left|F F^{\prime}\right| \sin \frac{\pi}{3}=\frac{21 \sqrt{3}}{4} .
\end{array}
$$