The area of the region $$\{(x, y) : 0 \leq y \leq 6 - x, y^2 \geq 4x - 3, x \geq 0\}$$ is:

$$6 - y = \frac{y^2 + 3}{4}$$

$$24 - 4y = y^2 + 3$$

$$y^2 + 4y - 21 = 0$$

$$(y + 7)(y - 3) = 0 \implies y = 3 \quad (\text{since } y \ge 0)$$

The point of intersection is $$(3, 3)$$

From $$y = 0$$ to $$y = 3$$: The region is bounded on the right by the parabola $$x = \frac{y^2 + 3}{4}$$.

From $$y = 3$$ to $$y = 6$$: The region is bounded on the right by the line $$x = 6 - y$$.

$$\text{Area} = \int_{0}^{3} \left( \frac{y^2 + 3}{4} \right) dy + \int_{3}^{6} (6 - y) \, dy$$

$$I_1 = \frac{1}{4} \left[ \frac{y^3}{3} + 3y \right]_{0}^{3} = \frac{1}{4} \left( \frac{27}{3} + 3(3) - 0 \right) = \frac{1}{4} (9 + 9) = \frac{18}{4} = \frac{9}{2}$$

$$I_2 = \int_{3}^{6} (6 - y) \, dy = \left[ 6y - \frac{y^2}{2} \right]_{3}^{6}$$

$$I_2 = \left( 36 - \frac{36}{2} \right) - \left( 18 - \frac{9}{2} \right) = 18 - \frac{27}{2} = \frac{9}{2}$$

$$\text{Total Area} = I_1 + I_2 = \frac{9}{2} + \frac{9}{2} = 9$$