12. (15 points) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram.
12. Given that $\left|F_{1} F_{2}\right|=2$.

As shown in Figure 5, the inscribed quadrilateral $\square A B C D$ of the ellipse has a pair of opposite sides $B C$ and $A D$ passing through the foci $F_{1}$ and $F_{2}$, respectively.
Obviously, the center of symmetry of $\square A B C D$ is the origin.
Figure 5

Let the points be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$.
Then $S_{\square A B C D}=2 S_{\triangle A F, D}=\left|F_{1} F_{2}\right|\left|y_{1}-y_{4}\right|$ $=2\left|y_{1}-y_{4}\right|$.
If the slope of line $A D$ exists, let its equation be
$$
y=k(x-1) \text {. }
$$

Substituting into the ellipse equation and simplifying, we get
$$
\left(3+4 k^{2}\right) x^{2}-8 k^{2} x+4 k^{2}-12=0 \text {. }
$$

By Vieta's formulas, we have
$$
x_{1}+x_{4}=\frac{8 k^{2}}{3+4 k^{2}}, x_{1} x_{4}=\frac{4 k^{2}-12}{3+4 k^{2}} \text {. }
$$

Then $\left|y_{1}-y_{4}\right|^{2}=k^{2}\left|x_{1}-x_{4}\right|^{2}$
$$
\begin{array}{l}
=k^{2}\left[\left(x_{1}+x_{4}\right)^{2}-4 x_{1} x_{4}\right] \\
=\frac{k^{2}}{\left(3+4 k^{2}\right)^{2}}\left[64 k^{4}-16\left(3+4 k^{2}\right)\left(k^{2}-3\right)\right] \\
=9 \times \frac{16 k^{2}\left(k^{2}+1\right)}{\left(3+4 k^{2}\right)^{2}} .
\end{array}
$$

Thus, $S_{\triangle \triangle B C D}=2\left|y_{1}-y_{4}\right|$
$$
=6 \sqrt{\frac{16 k^{2}\left(1+k^{2}\right)}{\left(3+4 k^{2}\right)^{2}}}<6 \text {. }
$$

If the slope of $A D$ does not exist, i.e., $A D \perp x$ axis.
It is easy to get the points $A\left(1, \frac{3}{2}\right), D\left(1,-\frac{3}{2}\right)$.
Thus, $S_{\text {ОАBCD }}=2\left|y_{1}-y_{4}\right|=6$.
In summary, the maximum area of $\square A B C D$ is 6.