Given, 

$\overrightarrow{a}=\alpha \hat{i}+2\hat{j}-\hat{k},\overrightarrow{b}=-2\hat{i}+\alpha \hat{j}+\hat{k}$,

Area of parallelogram $=\left|\hat{a}\times \hat{b}\right|$

$\hat{a}\times \hat{b}=\sqrt{{\left(\alpha +2\right)}^2+{\left(\alpha -2\right)}^2+{\left({\alpha }^2+4\right)}^2}$

Given $\left|\hat{a}\times \hat{b}\right|=\sqrt{15\left({\alpha }^2+4\right)}$

So, $2\left({\alpha }^2+4\right)+{\left({\alpha }^2+4\right)}^2=15\left({\alpha }^2+4\right)$

$\Rightarrow {\left({\alpha }^2+4\right)}^2=13\left({\alpha }^2+4\right)$

$\Rightarrow {\alpha }^2+4=13 \therefore {\alpha }^2=9$

Now, $2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}·\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2$

${\left|\overrightarrow{a}\right|}^2={\alpha }^2+4+1={\alpha }^2+5$

${\left|\overrightarrow{\mathrm{b}}\right|}^2=4+{\alpha }^2+1={\alpha }^2+5$

$\overrightarrow{a}·\overrightarrow{b}=-2\alpha +2\alpha -1=-1$

$\therefore 2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}·\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2$

$2\left({\alpha }^2+5\right)-1\left({\alpha }^2+5\right)={\alpha }^2+5=14$