Let $$f$$ and $$g$$ be twice differentiable functions on $$R$$ such that
$$f''(x) = g''(x) + 6x$$
$$f'(1) = 4g'(1) - 3 = 9$$
$$f(2) = 3, g(2) = 12$$
Then which of the following is NOT true?

Integrating $$f''(x) = g''(x) + 6x$$ gives $$f'(x) = g'(x) + 3x^2 + C_1$$. From $$f'(1) = 9$$ and $$4g'(1) - 3 = 9 \Rightarrow g'(1) = 3$$: $$9 = 3 + 3 + C_1$$, so $$C_1 = 3$$. Thus $$f'(x) - g'(x) = 3x^2 + 3$$.

Integrating again: $$f(x) - g(x) = x^3 + 3x + C_2$$. Using $$f(2) = 3$$ and $$g(2) = 12$$: $$3 - 12 = 8 + 6 + C_2$$, giving $$C_2 = -23$$. Hence $$f(x) - g(x) = x^3 + 3x - 23$$.

Option (B) states: if $$-1 < x < 2$$, then $$|f(x) - g(x)| < 8$$. Since $$h(x) = x^3 + 3x - 23$$ is strictly increasing ($$h'(x) = 3x^2 + 3 > 0$$), its values on $$(-1, 2)$$ range from $$h(-1) = -27$$ to $$h(2) = -9$$. So $$|f(x) - g(x)|$$ ranges from $$9$$ to $$27$$, which is always greater than $$8$$. Option (B) is NOT true.

The answer is $$\boxed{\text{Option (B)}}$$.