$\text{Let }S=\left\{\right(x,y)\in R\times R:x\ge 0,y\ge 0,y^2\le 4x,y^2\le 12-2x\text{ and}\mathrm{ 3}y+\sqrt{8x}\le 5\sqrt{8}\}.\text{If the area of the}$

$\text{region}\mathrm{ S }\text{is}\mathrm{ \alpha }\sqrt{2},\text{then}\mathrm{ \alpha }\text{ is equal to}$

The correct answer is 17/3.

We are given the set of points $(x,y)$ defining the region $S$ in the first quadrant ($x\ge 0,y\ge 0$) satisfying three key boundary conditions:

1. $y^2\le 4x\Rightarrow x\ge \frac{y^2}{4}$
2. $y^2\le 12-2x\Rightarrow x\le 6-\frac{y^2}{2}$
3. $3y+\sqrt{8}x\le 5\sqrt{8}\Rightarrow 2x+\frac{3}{\sqrt{2}}y\le 10\Rightarrow x\le 5-\frac{3y}{2\sqrt{2}}$

Step 1: Finding points of intersection

Let us find the point of intersection of the parabola $x=\frac{y^2}{4}$ and the line $x=5-\frac{3y}{2\sqrt{2}}$:

$\frac{y^2}{4}=5-\frac{3y}{2\sqrt{2}}$

Multiplying by 4:

$y^2+3\sqrt{2}y-20=0$

Factoring the quadratic equation:

$(y-2\sqrt{2})(y+5\sqrt{2})=0$

Since $y\ge 0$, we get $y=2\sqrt{2}$, which gives $x=\frac{(2\sqrt{2}{)}^2}{4}=2$.

Similarly, checking the point of intersection between the two parabolas $y^2=4x$ and $y^2=12-2x$:

$4x=12-2x\Rightarrow 6x=12\Rightarrow x=2,y=2\sqrt{2}$

Notice that the line $x=5-\frac{3y}{2\sqrt{2}}$ lies strictly inside the region bounded by $x=6-\frac{y^2}{2}$ for $y\in [0,2\sqrt{2}]$. Thus, the upper bound on $x$ for a given $y$ is dictated by the line.

Step 2: Setting up the Area Integral

For $y$ ranging from $0$ to $2\sqrt{2}$, $x$ varies from the left curve $x_{\text{left}}=\frac{y^2}{4}$ to the right boundary $x_{\text{right}}=5-\frac{3y}{2\sqrt{2}}$.

The total area of region $S$ is given by:

$\text{Area}={\int }_0^{2\sqrt{2}}\left(5-\frac{3y}{2\sqrt{2}}-\frac{y^2}{4}\right)dy$

Step 3: Evaluating the Integral

Integrating term-by-term:

$\text{Area}={\left[5y-\frac{3y^2}{4\sqrt{2}}-\frac{y^3}{12}\right]}_0^{2\sqrt{2}}$

Substituting the upper limit $y=2\sqrt{2}$:

$5\left(2\sqrt{2}\right)=10\sqrt{2}$
$\frac{3(2\sqrt{2}{)}^2}{4\sqrt{2}}=\frac{3\times 8}{4\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2}$
$\frac{(2\sqrt{2}{)}^3}{12}=\frac{16\sqrt{2}}{12}=\frac{4}{3}$

Subtracting these values:

$\text{Area}=10\sqrt{2}-3\sqrt{2}-\frac{4}{3}$

$\text{Area}=7\sqrt{2}-\frac{4}{3}$

Given that the area of the region is $\alpha \sqrt{2}$, by comparing both expressions, we find:

$\alpha =\frac{17}{3}$