Let the image of parabola $$x^{2}=4y$$, in the line x - y = 1 be $$(y+a)^{2}$$ = b(x-c), $$a,b,c \in N.$$ Then a + b + c is equal to

We are asked to find the image of the parabola $$x^2 = 4y$$ in the line $$x - y = 1$$ and express it in the form $$(y + a)^2 = b(x - c)$$, where $$a, b, c \in \mathbb{N}$$, then compute $$a + b + c$$.

For reflecting a point $$(h, k)$$ in the line $$x - y - 1 = 0$$, one uses the formulas $$\frac{h' - h}{1} = \frac{k' - k}{-1} = \frac{-2(h - k - 1)}{1^2 + (-1)^2}$$. From these, simplifying gives $$h' - h = -(h - k - 1) = k - h + 1$$, so $$h' = k + 1$$, and $$k' - k = (h - k - 1)$$, so $$k' = h - 1$$.

Rewriting these relations to express the original coordinates in terms of the reflected coordinates yields $$h = k' + 1$$ and $$k = h' - 1$$.

Since the original parabola satisfies $$h^2 = 4k$$, substituting the expressions for $$h$$ and $$k$$ gives $$(k' + 1)^2 = 4(h' - 1)$$. Renaming $$h'$$ as $$x$$ and $$k'$$ as $$y$$, this becomes $$(y + 1)^2 = 4(x - 1)$$.

Comparing $$(y + 1)^2 = 4(x - 1)$$ with the desired form $$(y + a)^2 = b(x - c)$$, we see that $$a = 1$$, $$b = 4$$, and $$c = 1$$. All of these are natural numbers, so $$a + b + c = 1 + 4 + 1 = 6$$.

The correct answer is Option (2): 6.