Let the area of the triangle with vertices $$A(1, \alpha)$$, $$B(\alpha, 0)$$ and $$C(0, \alpha)$$ be $$4$$ sq. units. If the points $$(\alpha, -\alpha)$$, $$(-\alpha, \alpha)$$ and $$(\alpha^2, \beta)$$ are collinear, then $$\beta$$ is equal to

We are given triangle with vertices $$A(1, \alpha)$$, $$B(\alpha, 0)$$, and $$C(0, \alpha)$$ with area 4 sq. units. Using the area formula Area $$= \frac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$ we compute $$ = \frac{1}{2}|1(0 - \alpha) + \alpha(\alpha - \alpha) + 0(\alpha - 0)| = \frac{1}{2}|-\alpha + 0 + 0| = \frac{|\alpha|}{2}. $$ Setting this equal to 4 gives $$\frac{|\alpha|}{2} = 4 \implies |\alpha| = 8 \implies \alpha = \pm 8.

The points $$($$\alpha$$, -$$\alpha$$)$$, $$(-$$\alpha$$, $$\alpha$$)$$, and $$($$\alpha^2$$, $$\beta$$)$$ are collinear. The slope between the first two points is $$m = $$\frac{\alpha - (-\alpha)}{-\alpha - \alpha} = \frac{2\alpha}{-2\alpha}$$ = -1,$$ and equating this to the slope between $$($$\alpha$$, -$$\alpha$$)$$ and $$($$\alpha^2$$, $$\beta$$)$$ gives $$\frac{\beta - (-\alpha)}{\alpha^2 - \alpha}$$ = -1.$$ Thus $$\beta + \alpha$$ = -($$\alpha^2 - \alpha$$) = -$$\alpha^2 + \alpha$$ and so $$\beta$$ = -$$\alpha^2$$.$$ For both $$\alpha$$ = 8$$ and $$\alpha$$ = -8$$, $$\beta$$ = -(8)^2 = -64 \quad $$\text{or}$$ \quad $$\beta$$ = -(-8)^2 = -64.

Therefore, $$\beta = $$ Option C: $$-64$$.