3. From the center of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, two perpendicular chords $A C$ and $B D$ are drawn. Connecting $A, B, C, D$ in sequence forms a quadrilateral. Then, the maximum value of the area $S$ of quadrilateral $A B C D$ is
3. 12.

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Given $O A \perp O B$, we have
$$
\begin{array}{l}
x_{1} x_{2}+y_{1} y_{2}=0 . \\
\begin{aligned}
\therefore S & =4 S_{\triangle A O B}=2|O A| \cdot|O B| \\
& =2 \sqrt{\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)} \\
& =2 \sqrt{\left(x_{1} x_{2}+y_{1} y_{2}\right)^{2}+\left(x_{1} y_{2}-x_{2} y_{1}\right)^{2}} \\
& =2 \sqrt{\left(x_{1} y_{2}-x_{2} y_{1}\right)^{2}} \\
& =12 \sqrt{\left(\frac{x_{1}}{3} \cdot \frac{y_{2}}{2}-\frac{x_{2}}{3} \cdot \frac{y_{1}}{2}\right)^{2}} \\
& =12 \sqrt{\left(\frac{x_{1}^{2}}{9}+\frac{y_{1}^{2}}{4}\right)\left(\frac{x_{2}^{2}}{9}+\frac{y_{2}^{2}}{4}\right)-\left(\frac{x_{1} x_{2}}{9}+\frac{y_{1} y_{2}}{4}\right)^{2}} \\
& =12 \sqrt{1-\left(\frac{x_{1} x_{2}}{9}+\frac{y_{1} y_{2}}{4}\right)^{2}} \leqslant 12 .
\end{aligned}
\end{array}
$$

Equality holds if and only if $\frac{x_{1} x_{2}}{9}+\frac{y_{1} y_{2}}{4}=0$. Given $x_{1} x_{2}+$ $y_{1} y_{2}=0$, it follows that $x_{1} x_{2}=0, y_{1} y_{2}=0$. In this case, points $A$ and $B$ are the endpoints of the major and minor axes of the ellipse, respectively.