Let $S=\{(x,y)\in <!--\; \in \; -->\mathbb{R}\times <!--\; \times \; -->\mathbb{R}:x\ge <!--\; \ge \; -->0,y\ge <!--\; \ge \; -->0,$ y^2\le <!--\; \le \; -->4x,y^2\le <!--\; \le \; -->12-<!--\; -\; -->2x$$ and $3y+\sqrt{8}x\le <!--\; \le \; -->5\sqrt{8}\}$. If the area of the region S is α√2, then α is equal to

To find the area of the region $S$, we first analyze the given boundaries:
1. $x\ge 0$ and $y\ge 0$ (which restricts the region to the first quadrant).
2. $y^2\le 4x\Rightarrow x\ge \frac{y^2}{4}$
3. $y^2\le 12-2x\Rightarrow x\le 6-\frac{y^2}{2}$
4. $3y+\sqrt{8}x\le 5\sqrt{8}\Rightarrow x\le 5-\frac{3}{\sqrt{8}}y$

First, let us find the intersection points of these curves:
For the two parabolas, $x=\frac{y^2}{4}$ and $x=6-\frac{y^2}{2}$:
$\frac{y^2}{4}=6-\frac{y^2}{2}$
Multiplying by 4:
$y^2=24-2y^2\Rightarrow 3y^2=24\Rightarrow y^2=8$
Since $y\ge 0$, we have $y=\sqrt{8}=2\sqrt{2}$.
Substituting $y=\sqrt{8}$ into $x=\frac{y^2}{4}$ gives:
$x=\frac{8}{4}=2$
So, the intersection point of the two parabolas is $(2,\sqrt{8})$.

Now, let us check where the boundary line $3y+\sqrt{8}x=5\sqrt{8}$ passes.
If we substitute $y=\sqrt{8}$ and $x=2$ into the equation:
$3(\sqrt{8})+\sqrt{8}(2)=5\sqrt{8}$
This satisfies the equation, showing that all three boundary curves intersect at the single point $(2,\sqrt{8})$.

Let us determine the bounds of $x$ in terms of $y$ for the region $S$:
For $0\le y\le \sqrt{8}$, we have:
$\frac{y^2}{4}\le x\le 5-\frac{3}{\sqrt{8}}y$
Therefore, the area $A$ of the region $S$ is given by the integral:
$A={\int }_0^{\sqrt{8}}\left(5-\frac{3}{\sqrt{8}}y-\frac{y^2}{4}\right)dy$

Integrating term by term:
$A={\left[5y-\frac{3}{2\sqrt{8}}\right]}_0^{\sqrt{8}}$

Evaluating this at the upper limit $y=\sqrt{8}$:
1. First term: $5\sqrt{8}=10\sqrt{2}$
2. Second term: $-\frac{3}{2\sqrt{8}}(8)=-\frac{12}{\sqrt{8}}=-\frac{12}{2\sqrt{2}}=-3\sqrt{2}$
3. Third term: $-\frac{{(\sqrt{8})}^3}{12}=-\frac{8\sqrt{8}}{12}=-\frac{2}{3}(2\sqrt{2})=-\frac{4}{3}$

Combining the terms:
$A=10\sqrt{2}-3\sqrt{2}-\frac{4}{3}$
$A=7\sqrt{2}-\frac{4}{3}$
$A=\left(7-\frac{4}{3}\right)\sqrt{2}=\frac{17}{3}$

Given that the area of region $S$ is $\alpha \sqrt{2}$, we compare it with our result:
$\alpha \sqrt{2}=\frac{17}{3}$

Thus, the correct option is 17/3.