If $$f(x) = x^3 - x^2f'(1) + xf''(2) - f'''(3)$$, $$x \in \mathbb{R}$$, then

We are given $$f(x) = x^3 - x^2 f'(1) + x f''(2) - f'''(3)$$ for $$x \in \mathbb{R}$$. Setting $$a=f'(1)$$, $$b=f''(2)$$, and $$c=f'''(3)$$ gives $$f(x)=x^3-ax^2+bx-c$$.

Computing the first derivative yields $$f'(x)=3x^2-2ax+b$$. Plugging in $$x=1$$, we get $$f'(1)=3-2a+b$$, which must equal $$a$$ and leads to $$3-2a+b=a$$ or $$3+b=3a$$. The second derivative is $$f''(x)=6x-2a$$, so $$f''(2)=12-2a=b$$. Finally, the third derivative is constant: $$f'''(x)=6$$, hence $$c=6$$.

Using $$b=12-2a$$ in the equation $$3+b=3a$$ gives $$3+12-2a=3a$$, which simplifies to $$15=5a$$ and thus $$a=3$$. It follows that $$b=12-6=6$$ and $$c=6$$. Therefore $$f(x)=x^3-3x^2+6x-6$$.

Evaluating at specific points gives:
$$f(0)=-6$$
$$f(1)=1-3+6-6=-2$$
$$f(2)=8-12+12-6=2$$
$$f(3)=27-27+18-6=12$$.

Checking the options:
Option A: $$3f(1)+f(2)=3(-2)+2=-4\neq12=f(3)$$ ✘
Option B: $$f(3)-f(2)=12-2=10\neq-2=f(1)$$ ✘
Option C: $$2f(0)-f(1)+f(3)=2(-6)-(-2)+12=-12+2+12=2=f(2)$$ ✔
Option D: $$f(1)+f(2)+f(3)=-2+2+12=12\neq-6=f(0)$$ ✘

The correct answer is Option C: $$2f(0)-f(1)+f(3)=f(2)$$.