The maximum area of such a parallelogram $A F D E$, with one vertex fixed at $A$ and the other three points lying on the sides of triangle $A B C$, is half the area of triangle $A B C$.

Using the determinant formula for area of triangle with vertices $A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right)$ :

Area $\triangle A B C$

$$=\frac{1}{2}\left|x_1\left(y_2-y_3^2\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$$

Substitute the coordinates:

$$\begin{aligned} & =\frac{1}{2}|4(1-(-3))+1((-3)-(-2))+9((-2)-1)| \\ & =\frac{1}{2}|4(4)+1(-1)+9(-3)| \\ & =\frac{1}{2}|16-1-27|=\frac{1}{2}|-12|=\frac{12}{2}=6 \end{aligned}$$

Maximum area of parallelogram $A F D E$ $=\frac{1}{2} \times$ area of triangle $=\frac{1}{2} \times 6=3$