Let $A(4, -2)$, $B(1, 1)$ and $C(9, -3)$ be the vertices of a triangle $ABC$. Then the maximum area of the parallelogram $AFDE$, formed with vertices $D$, $E$, and $F$ on the sides $BC$, $CA$ and $AB$ of the triangle $ABC$ respectively, is:

Step 1: Calculate the area of triangle ABC

To find the area of the parallelogram, we first need to calculate the area of the triangle $ABC$. We use the determinant formula for the area of a triangle given its vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$.

Step 2: Substitute coordinates and compute area

Substitute the coordinates $A(4, -2)$, $B(1, 1)$, and $C(9, -3)$ into the formula. After performing the arithmetic operations, we find the area of triangle $ABC$ to be 6 square units.

Step 3: Relate parallelogram area to triangle area

For a parallelogram $AFDE$ inscribed in a triangle $ABC$ with vertices $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively, the maximum area of the parallelogram is half the area of the triangle. This is a known geometric property.

Step 4: Calculate maximum area of parallelogram

Using the property that the maximum area of the inscribed parallelogram is half the area of the triangle, we multiply the calculated area of triangle $ABC$ by $\frac{1}{2}$ to find the maximum area of parallelogram $AFDE$.

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