If the curve $$y = ax^2 + bx + c$$, $$x \in R$$, passes through the point (1, 2) and the tangent line to this curve at origin is $$y = x$$, then the possible values of $$a, b, c$$ are:

The curve $$y = ax^2 + bx + c$$ passes through the point $$(1, 2)$$ and has its tangent at the origin equal to $$y = x$$.

Since the tangent line at the origin is $$y = x$$, the curve must pass through the origin. Setting $$x = 0$$ and $$y = 0$$, we get $$0 = a(0) + b(0) + c$$, so $$c = 0$$.

The slope of the tangent at the origin must equal the slope of $$y = x$$, which is $$1$$. Differentiating, $$\frac{dy}{dx} = 2ax + b$$. At $$x = 0$$, this gives $$b = 1$$.

Since the curve passes through $$(1, 2)$$, we have $$2 = a(1)^2 + b(1) + c = a + 1 + 0$$, so $$a = 1$$.

Therefore, the values are $$a = 1, b = 1, c = 0$$.