Now $a c-b^2=2$ and $2 a+b=1$ and $2 b + c =2$

solving all these above equations we get

$\frac{1-b}{2} \times\left(\frac{2-2 b}{1}\right)-b^2=2$

$\Rightarrow(1-b)^2-b^2=2$

$\Rightarrow 1-2 b=2$

$\Rightarrow b=-\frac{1}{2} \text { and } a=\frac{3}{4} \text { and } c=3$

Hence $\alpha=3 a+\frac{3 b}{2}=\frac{9}{4}-\frac{3}{4}=\frac{3}{2}$

and $\beta=3 b+\frac{3 c}{2}=-\frac{3}{2}+\frac{9}{2}=3$

also $s = a + c =\frac{15}{4}$

$\therefore \frac{\beta s}{\alpha^2}=\frac{3 \times 15}{4 \times \frac{9}{4}}=5$