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 __________.

Let us place the origin at vertex $$A(4,-2)$$ and write every point as
$$\text{position} = A + \text{vector}$$.

Vectors along the two sides through $$A$$ are
$$\mathbf{b} = \overrightarrow{AB} = B-A = (1-4,\,1-(-2)) = (-3,\,3)$$
$$\mathbf{c} = \overrightarrow{AC} = C-A = (9-4,\,-3-(-2)) = (5,\,-1).$$

Take the points on the three sides as
$$F = A + s\,\mathbf{b},\; 0\le s\le 1$$ (on $$AB$$)
$$E = A + u\,\mathbf{c},\; 0\le u\le 1$$ (on $$AC$$)
$$D = B + v\,(C-B),\; 0\le v\le 1$$ (on $$BC$$).

The vertices of the required parallelogram are in the order $$A\!-\!F\!-\!D\!-\!E$$, so the adjacent edges are
$$\overrightarrow{AF} = s\,\mathbf{b},$$
$$\overrightarrow{AE} = u\,\mathbf{c}.$$
For a parallelogram we must also have
$$\overrightarrow{AE} = \overrightarrow{FD}.$$\

First compute $$\overrightarrow{FD}$$:
$$F = (4-3s,\,-2+3s),$$
$$D = \bigl(1+8v,\,1-4v\bigr),$$
hence $$\overrightarrow{FD}=D-F=\bigl(-3+8v+3s,\,3-4v-3s\bigr).$$

Setting $$\overrightarrow{AE}=u\,\mathbf{c}=(5u,\,-u)$$ equal to $$\overrightarrow{FD}$$ gives the system
$$5u = -3 + 8v + 3s\quad -(1)$$
$$-u = 3 - 4v - 3s\quad -(2).$$

Multiply $$(2)$$ by $$-5$$ and add to $$(1)$$:
$$5u = -3 + 8v + 3s,$$
$$5u = 15 - 20v - 15s,$$
leading to $$12 - 12v - 12s = 0 \;\Longrightarrow\; s + v = 1.$$

Substituting $$s = 1 - v$$ in $$(1)$$ or $$(2)$$ yields $$u = v.$$\

Thus the parameters are constrained by
$$s = 1 - v,\quad u = v,\quad 0 \le v \le 1.$$

The area of a parallelogram formed by adjacent side vectors $$\mathbf{p}$$ and $$\mathbf{q}$$ is $$|\mathbf{p}\times\mathbf{q}|$$ (magnitude of the 2-D cross product).
Here
$$\mathbf{p} = \overrightarrow{AF} = s\,\mathbf{b},\quad \mathbf{q} = \overrightarrow{AE} = u\,\mathbf{c} = v\,\mathbf{c}.$$
Hence
$$\text{Area} = |s\,\mathbf{b}\times v\,\mathbf{c}| = sv\,|\mathbf{b}\times\mathbf{c}|.$$\

Compute $$|\mathbf{b}\times\mathbf{c}|$$:
$$\mathbf{b}\times\mathbf{c} = \begin{vmatrix} -3 & 3 \\ 5 & -1 \end{vmatrix} = (-3)(-1) - 3(5) = 3 - 15 = -12,$$
so $$|\mathbf{b}\times\mathbf{c}| = 12.$$

Therefore
$$\text{Area} = 12\,sv = 12\,v(1-v)$$ because $$s = 1 - v.$$\

The quadratic $$v(1-v) = v - v^{2}$$ attains its maximum at
$$\frac{d}{dv}[v - v^{2}] = 1 - 2v = 0 \;\Longrightarrow\; v = \frac12,$$
with the same $$s = 1 - v = \dfrac12.$$

Maximum area
$$= 12\left(\frac12\right)\left(\frac12\right) = 3.$$

Hence the greatest possible area of the parallelogram $$AFDE$$ is $$\boxed{3}$$.