Let the area of the region $$\{(x, y) : 0 \leq x \leq 3, 0 \leq y \leq \min\{x^2 + 2, 2x + 2\}\}$$ be $$A$$. Then $$12A$$ is equal to ______.

We need to find the area of the region $$\{(x, y) : 0 \leq x \leq 3, \, 0 \leq y \leq \min\{x^2 + 2, \, 2x + 2\}\}$$.

Find where the curves intersect: $$x^2 + 2 = 2x + 2 \implies x^2 - 2x = 0 \implies x(x - 2) = 0$$

So the curves intersect at $$x = 0$$ and $$x = 2$$.

Determine which function is smaller on each interval: for $$0 < x < 2$$: $$x^2 + 2 < 2x + 2$$ (since $$x^2 < 2x$$ for $$0 < x < 2$$).

For $$x > 2$$: $$x^2 + 2 > 2x + 2$$.

So $$\min = x^2 + 2$$ on $$[0, 2]$$ and $$\min = 2x + 2$$ on $$[2, 3]$$.

Compute the area: $$ A = \int_0^2 (x^2 + 2) \, dx + \int_2^3 (2x + 2) \, dx $$

First integral:

$$\int_0^2 (x^2 + 2) \, dx = \left[\frac{x^3}{3} + 2x\right]_0^2 = \frac{8}{3} + 4 = \frac{20}{3}$$

Second integral:

$$\int_2^3 (2x + 2) \, dx = [x^2 + 2x]_2^3 = (9 + 6) - (4 + 4) = 15 - 8 = 7$$

$$ A = \frac{20}{3} + 7 = \frac{20 + 21}{3} = \frac{41}{3} $$

Compute $$12A$$: $$ 12A = 12 \times \frac{41}{3} = 4 \times 41 = 164 $$

The answer is $$\boxed{164}$$.