The problem can be represented as follows, where $AB$ is the latus rectum of the parabola $x^2 = 12y$.

Given $x^2 =12y \rightarrow 4a = 12 \rightarrow a = 3$.

Therefore, the coordinates of point $C = (0,3)$

When $y = 3 \rightarrow x^2 = 12 \times 3 = 36 \rightarrow x = \pm 6$

Therefore $A = (-6,3)$ and $B = (6,3)$.

Given the 3 vertices of the triangle, we can calculate the area as follows: $\text{Area} = |\frac12(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))|$

Now that we know the vertices of $\Delta AOB$, we can calculate the area.

$\qquad = |\frac12( 0 (3-3) + (-6) (3-0) + 6(0-3)|$

$\qquad = |\frac12(-18-18)|$

$\qquad = \large\frac{1}{2}$$\times 36 = 18\; \text{sq. units}$