The sum of all the local minimum values of the twice differentiable function $$f : R \to R$$ defined by $$f(x) = x^3 - 3x^2 - \frac{3f''(2)}{2}x + f''(1)$$ is:

Given $$f(x) = x^3 - 3x^2 - \frac{3f''(2)}{2}x + f''(1)$$, we first determine $$f''(2)$$ and $$f''(1)$$.

Differentiating twice: $$f'(x) = 3x^2 - 6x - \frac{3f''(2)}{2}$$ and $$f''(x) = 6x - 6$$.

Evaluating: $$f''(2) = 12 - 6 = 6$$ and $$f''(1) = 6 - 6 = 0$$.

Substituting back: $$f(x) = x^3 - 3x^2 - \frac{3 \cdot 6}{2}x + 0 = x^3 - 3x^2 - 9x.$$

Critical points: $$f'(x) = 3x^2 - 6x - 9 = 3(x-3)(x+1) = 0$$, giving $$x = 3$$ and $$x = -1$$.

Using the second derivative test: $$f''(3) = 12 > 0$$ (local minimum) and $$f''(-1) = -12 < 0$$ (local maximum).

The only local minimum value is $$f(3) = 27 - 27 - 27 = -27$$.

The sum of all local minimum values is $$-27$$.