$$\because$$ A(1, $$\alpha$$), B($$\alpha$$, 0) and C(0, $$\alpha$$) are the vertices of $$\Delta$$ABC and area of $$\Delta$$ABC = 4

$$\therefore$$ $$\left| {{1 \over 2}\left| {\matrix{ 1 & \alpha & 1 \cr \alpha & 0 & 1 \cr 0 & \alpha & 1 \cr } } \right|} \right| = 4$$

$$ \Rightarrow \left| {1(1 - \alpha ) - \alpha (\alpha ) + {\alpha ^2}} \right| = 8$$

$$ \Rightarrow \alpha = \, \pm \,8$$

Now, $$(\alpha ,\, - \alpha ),\,( - \alpha ,\alpha )$$ and $$({\alpha ^2},\beta )$$ are collinear

$$\therefore$$ $$\left| {\matrix{ 8 & { - 8} & 1 \cr { - 8} & 8 & 1 \cr {64} & \beta & 1 \cr } } \right| = 0 = \left| {\matrix{ { - 8} & 8 & 1 \cr 8 & { - 8} & 1 \cr {64} & \beta & 1 \cr } } \right|$$

$$ \Rightarrow 8(8 - \beta ) + 8( - 8 - 64) + 1( - 8\beta - 8 \times 64) = 0$$

$$ \Rightarrow 8 - \beta - 72 - \beta - 64 = 0$$

$$ \Rightarrow \beta = - 64$$