The value of $$\int_{-2}^{2} |3x^2 - 3x - 6| dx$$ is ______

We factor the expression inside the absolute value: $$3x^2 - 3x - 6 = 3(x^2 - x - 2) = 3(x - 2)(x + 1)$$.

The roots are $$x = -1$$ and $$x = 2$$. On $$[-2, -1]$$: $$(x-2) < 0$$ and $$(x+1) \leq 0$$, so the product is non-negative, and $$|3(x-2)(x+1)| = 3(x^2 - x - 2)$$. On $$[-1, 2]$$: $$(x-2) \leq 0$$ and $$(x+1) \geq 0$$, so the product is non-positive, and $$|3(x-2)(x+1)| = -3(x^2 - x - 2) = 3(2 + x - x^2)$$.

Computing the first integral: $$\int_{-2}^{-1} 3(x^2 - x - 2)\,dx = 3\left[\frac{x^3}{3} - \frac{x^2}{2} - 2x\right]_{-2}^{-1}$$.

At $$x = -1$$: $$\frac{-1}{3} - \frac{1}{2} + 2 = \frac{-2 - 3 + 12}{6} = \frac{7}{6}$$. At $$x = -2$$: $$\frac{-8}{3} - \frac{4}{2} + 4 = \frac{-8}{3} - 2 + 4 = \frac{-8}{3} + 2 = \frac{-2}{3}$$.

So the first integral is $$3\left(\frac{7}{6} - \left(-\frac{2}{3}\right)\right) = 3\left(\frac{7}{6} + \frac{4}{6}\right) = 3 \times \frac{11}{6} = \frac{11}{2}$$.

Computing the second integral: $$\int_{-1}^{2} 3(2 + x - x^2)\,dx = 3\left[2x + \frac{x^2}{2} - \frac{x^3}{3}\right]_{-1}^{2}$$.

At $$x = 2$$: $$4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$$. At $$x = -1$$: $$-2 + \frac{1}{2} + \frac{1}{3} = \frac{-12 + 3 + 2}{6} = \frac{-7}{6}$$.

So the second integral is $$3\left(\frac{10}{3} - \left(-\frac{7}{6}\right)\right) = 3\left(\frac{20}{6} + \frac{7}{6}\right) = 3 \times \frac{27}{6} = \frac{27}{2}$$.

The total value is $$\frac{11}{2} + \frac{27}{2} = \frac{38}{2} = 19$$.