The maximum area of a triangle whose one vertex is at $$(0, 0)$$ and the other two vertices lie on the curve $$y = -2x^2 + 54$$ at points $$(x, y)$$ and $$(-x, y)$$ where $$y > 0$$ is :

One vertex is at the origin $$(0,0)$$, and the other two are at $$(x, y)$$ and $$(-x, y)$$ on the curve $$y = -2x^2 + 54$$ where $$y > 0$$.

The base of the triangle = $$2x$$, and the height = $$y$$.

Area = $$\frac{1}{2} \times 2x \times y = xy = x(-2x^2 + 54) = -2x^3 + 54x$$.

To maximize, take the derivative and set it to zero:

$$ \frac{dA}{dx} = -6x^2 + 54 = 0 \Rightarrow x^2 = 9 \Rightarrow x = 3 $$

(taking positive value since $$x > 0$$)

$$ y = -2(9) + 54 = -18 + 54 = 36 $$

Maximum area = $$3 \times 36 = 108$$.

Verify it's a maximum: $$\frac{d^2A}{dx^2} = -12x = -36 < 0$$. Confirmed maximum.

The answer is Option (4): $$\boxed{108}$$.