The two parabolas $y^2=4x$ and $y^2=12-2x$ meet where $4x=12-2x$, giving $x=2$ and $y=2\sqrt{2}$. So $A=(2,2\sqrt{2})$.

The line $3y+\sqrt{8}x=5\sqrt{8}$ passes through $A(2,2\sqrt{2})$ (check: $6\sqrt{2}+2\sqrt{8}=6\sqrt{2}+4\sqrt{2}=10\sqrt{2}=5\sqrt{8}$) and hits the x-axis at $B(5,0)$. Let $C=(2,0)$.

Split $S$ into two pieces:

(i) Region under $y=2\sqrt{x}$ from $x=0$ to $x=2$:

${\int }_0^{2}2\sqrt{x}dx={\left[\frac{4}{3}{x}^{3/2}\right]}_0^2=\frac{4}{3}\cdot 2\sqrt{2}=\frac{8\sqrt{2}}{3}.$

(ii) Triangle $ACB$ with base $CB=3$ and height $2\sqrt{2}$:

$\frac{1}{2}\cdot 3\cdot 2\sqrt{2}=3\sqrt{2}.$

Total area $=\frac{8\sqrt{2}}{3}+3\sqrt{2}=\frac{8\sqrt{2}+9\sqrt{2}}{3}=\frac{17\sqrt{2}}{3}=\alpha \sqrt{2}.$

Hence $\alpha =\frac{17}{3}$. Option (B) is correct.