If $$f(x) = x^2 + g'(1)x + g''(2)$$ and $$g(x) = f(1)x^2 + xf'(x) + f''(x)$$, then the value of $$f(4) - g(4)$$ is equal to _____.

We are given $$f(x) = x^2 + g'(1)x + g''(2)$$ and $$g(x) = f(1)x^2 + xf'(x) + f''(x)$$. We need to find $$f(4) - g(4)$$.

Let $$g'(1) = a$$ and $$g''(2) = b$$. Then $$f(x) = x^2 + ax + b$$. Since $$f'(x) = 2x + a$$ and $$f''(x) = 2$$, it follows that $$f(1) = 1 + a + b$$.

Substituting these into $$g(x) = f(1)x^2 + xf'(x) + f''(x)$$ gives $$g(x) = (1 + a + b)x^2 + x(2x + a) + 2 = (3 + a + b)x^2 + ax + 2$$.

Now differentiating yields $$g'(x) = 2(3 + a + b)x + a$$, so $$g'(1) = 6 + 3a + 2b$$. Since $$g'(1) = a$$, one obtains $$a = 6 + 3a + 2b$$, which simplifies to $$a + b = -3$$. Similarly, $$g''(x) = 2(3 + a + b)$$ implies $$g''(2) = 2(3 + a + b)$$, and because $$g''(2) = b$$, we get $$b = 6 + 2a + 2b$$, leading to $$2a + b = -6$$.

Solving the system $$a + b = -3$$ and $$2a + b = -6$$ gives $$a = -3$$ and $$b = 0$$.

Substituting these values back gives $$f(x) = x^2 - 3x + 0 = x^2 - 3x$$ and $$f(4) = 16 - 12 = 4$$. Moreover, $$g(x) = (3 + (-3) + 0)x^2 + (-3)x + 2 = 0 \cdot x^2 - 3x + 2 = -3x + 2$$ and thus $$g(4) = -12 + 2 = -10$$. Therefore, $$f(4) - g(4) = 4 - (-10) = 14$$.

The answer is $$\boxed{14}$$.