Let $\overrightarrow{\mathrm{a}}=\mathrm{\alpha }\stackrel{⏜}{\mathrm{i}}+2\stackrel{⏜}{\mathrm{j}}-\stackrel{⏜}{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{b}}=-2\stackrel{⏜}{\mathrm{i}}+\mathrm{\alpha }\stackrel{⏜}{\mathrm{j}}+\stackrel{⏜}{\mathrm{k}}$, where $\mathrm{\alpha }\in \mathrm{R}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\overrightarrow{\mathrm{a}}\text{ and }\overrightarrow{\mathrm{b}}\text{ is }\sqrt{15\left({\mathrm{\alpha }}^2+4\right)}$, then the value of $2|\overrightarrow{\mathrm{a}}{|}^2+\left(\begin{array}{c}\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}\end{array}\right)|\overrightarrow{\mathrm{b}}{|}^2$ is equal to :

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

area of parallelogram$=|\hat{\mathrm{a}}\times \hat{\mathrm{b}}|$

$$\begin{gathered} |\hat{\mathrm{a}}\times \hat{\mathrm{b}}|=\left|\begin{array}{ccc}i& j& k\\ \alpha & 2& -1\\ -2& \alpha & 1\end{array}\right| \\ =\sqrt{(\mathrm{\alpha }+2{)}^2+(\mathrm{\alpha }-2{)}^2+{\left({\mathrm{\alpha }}^2+4\right)}^2} \end{gathered}$$

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

$\begin{array}{l}2\left({\mathrm{\alpha }}^2+4\right)+{\left({\mathrm{\alpha }}^2+4\right)}^2=15\left({\mathrm{\alpha }}^2+4\right)\\ {\left({\mathrm{\alpha }}^2+4\right)}^2=13\left({\mathrm{\alpha }}^2+4\right)\\ \Rightarrow {\mathrm{\alpha }}^2+4=13\therefore {\mathrm{\alpha }}^2=9\\ 2|\overrightarrow{\mathrm{a}}{|}^2+(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}\left)\right|\overrightarrow{\mathrm{b}}{|}^2\\ |\overrightarrow{\mathrm{a}}{|}^2={\mathrm{\alpha }}^2+4+1={\mathrm{\alpha }}^2+5\\ |\overrightarrow{\mathrm{b}}{|}^2=4+{\mathrm{\alpha }}^2+1={\mathrm{\alpha }}^2+5\\ \overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=-2\mathrm{\alpha }+2\mathrm{\alpha }-1=-1\\ \therefore 2|\overrightarrow{\mathrm{a}}{|}^2+(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}\left)\right|\overrightarrow{\mathrm{b}}{|}^2\\ 2\left({\mathrm{\alpha }}^2+5\right)-1\left({\mathrm{\alpha }}^2+5\right)={\mathrm{\alpha }}^2+5=14\end{array}$