If the chord joining the points $$ P_{1}(x_{1}, y_{1}) $$ and $$P_{2}(x_{2},y_{2})$$ on the parabola $$y^{2}=12x$$ subtends a right angle at the vertex of the parabola, then $$ x_{1}x_{2}-y_{1}y_{2} $$ is equal to

Since the given parabola $$y^2 = 12x$$ can be written in the form $$y^2 = 4ax$$, we identify $$a = 3$$. Consequently, the parametric coordinates of the points are $$P_1 = (3t_1^2, 6t_1)$$ and $$P_2 = (3t_2^2, 6t_2)$$.

The vertex is at the origin $$(0, 0)$$. Because $$OP_1 \perp OP_2$$, the product of the slopes of $$OP_1$$ and $$OP_2$$ must satisfy:

$$\text{slope of } OP_1 \times \text{slope of } OP_2 = -1$$

$$\frac{6t_1}{3t_1^2} \times \frac{6t_2}{3t_2^2} = -1$$

$$\frac{2}{t_1} \times \frac{2}{t_2} = -1$$

This gives $$t_1 t_2 = -4$$.

We now compute the value of $$x_1 x_2 - y_1 y_2$$.

$$x_1 x_2 = 3t_1^2 \cdot 3t_2^2 = 9(t_1 t_2)^2 = 9 \times 16 = 144$$

$$y_1 y_2 = 6t_1 \cdot 6t_2 = 36 t_1 t_2 = 36 \times (-4) = -144$$

$$x_1 x_2 - y_1 y_2 = 144 - (-144) = 288$$