12. Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with left and right foci $F_{1}$ and $F_{2}$, respectively, a line $l$ passing through the right focus $F_{2}$ intersects the ellipse at points $P$ and $Q$. Then the maximum value of the area of the incircle of $\triangle F_{1} P Q$ is $\qquad$
12. $\frac{9 \pi}{16}$.

Let the equation of line $l$ be $x=m y+1, P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}x=m y+1, \\ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1,\end{array}\right.$ we get
$\left(3 m^{2}+4\right) y^{2}+6 m y-9=0$.
Thus, $y_{1}+y_{2}=\frac{-6 m}{3 m^{2}+4}, y_{1} y_{2}=-\frac{9}{3 m^{2}+4}$.
And $S_{\triangle F_{1} P Q}=\frac{1}{2} \times 2\left|y_{1}-y_{2}\right|$
$=\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}}=12 \sqrt{\frac{m^{2}+1}{\left(3 m^{2}+4\right)^{2}}}$.
Let $t=m^{2}+1$, then
$$
\sqrt{\frac{m^{2}+1}{\left(3 m^{2}+4\right)^{2}}}=\sqrt{\frac{t}{9 t^{2}+6 t+1}}=\sqrt{\frac{1}{9 t+\frac{1}{t}+6}} \text {. }
$$

The function $u=9 t+\frac{1}{t}(t \geqslant 1)$ is an increasing function, so when $t=1, m=0$, $u$ has its minimum value, at which point, the area is maximized. Since the perimeter of $\triangle F_{1} P Q$ is a constant 8, the inradius of $\triangle F_{1} P Q$ is maximized at this point. Therefore, its area is also maximized. Let the inradius of $\triangle F_{1} P Q$ at this point be $r$, then $\frac{1}{2} \times 8 r=3, r=\frac{3}{4}$. Thus, the maximum area of the incircle of $\triangle F_{1} P Q$ is $\frac{9 \pi}{16}$.