The area of the region \ (x, y) : 0 ≤ y ≤ 6 - x, y^2 ≥ 4x - 3, x ≥...

The given region is defined by the inequalities:
$x \geq 0$
$0 \leq y \leq 6 - x \Rightarrow y \geq 0$ and $x \leq 6 - y$
$y^2 \geq 4x - 3 \Rightarrow x \leq \dfrac{y^2 + 3}{4}$

From these inequalities, for a given $y \geq 0$, the value of $x$ ranges from $0$ to $\min\left(6 - y, \dfrac{y^2 + 3}{4}\right)$.

To find the point where the two bounding curves intersect, we equate them:
$6 - y = \dfrac{y^2 + 3}{4}$

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

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

$(y + 7)(y - 3) = 0$

Since $y \geq 0$, the intersection occurs at $y = 3$.

For $0 \leq y \leq 3$, the right boundary is the parabola $x = \dfrac{y^2 + 3}{4}$.
For $3 \leq y \leq 6$, the right boundary is the line $x = 6 - y$.

The total area $A$ can be calculated by integrating with respect to $y$:
$A = \int_{0}^{3} \dfrac{y^2 + 3}{4} dy + \int_{3}^{6} (6 - y) dy$

Evaluating the first integral:
$\int_{0}^{3} \dfrac{y^2 + 3}{4} dy = \dfrac{1}{4} \left[ \dfrac{y^3}{3} + 3y \right]_{0}^{3} = \dfrac{1}{4} (9 + 9) = \dfrac{18}{4} = \dfrac{9}{2}$

Evaluating the second integral:
$\int_{3}^{6} (6 - y) dy = \left[ 6y - \dfrac{y^2}{2} \right]_{3}^{6} = (36 - 18) - \left(18 - \dfrac{9}{2}\right) = 18 - 13.5 = \dfrac{9}{2}$

Total Area = $\dfrac{9}{2} + \dfrac{9}{2} = 9$

Answer: $9$