The parabolas: $$ax^2 + 2bx + cy = 0$$ and $$d^2 + 2ex + fy = 0$$ intersect on the line $$y = 1$$. If $$a, b, c, d, e, f$$ are positive real numbers and $$a, b, c$$ are in G.P., then

Given parabolas: $$ax^2 + 2bx + cy = 0$$ and $$dx^2 + 2ex + fy = 0$$ intersect on the line $$y = 1$$.

Note: The second equation appears to have a typo ($$d^2$$ instead of $$dx^2$$). Assuming it is $$dx^2 + 2ex + fy = 0$$.

Substituting $$y = 1$$ into the first equation gives $$ax^2 + 2bx + c = 0 \quad \ldots (i)$$.

Substituting $$y = 1$$ into the second equation gives $$dx^2 + 2ex + f = 0 \quad \ldots (ii)$$.

Since the parabolas intersect on $$y = 1$$, both equations must have the same roots and hence are proportional: $$\frac{a}{d} = \frac{2b}{2e} = \frac{c}{f}$$.

This simplifies to $$\frac{a}{d} = \frac{b}{e} = \frac{c}{f} = k \text{ (say)}$$.

Since $$a, b, c$$ are in G.P., we have $$b^2 = ac$$.

From the proportionality relations, $$d = a/k$$, $$e = b/k$$, $$f = c/k$$.

Then $$e^2 = \frac{b^2}{k^2} = \frac{ac}{k^2} = \frac{a}{k}\cdot\frac{c}{k} = d\cdot f$$, so $$d, e, f$$ are in G.P.

The correct answer is Option (4): $$\boxed{d, e, f \text{ are in G.P.}}$$