Let $\overline{a}and\overline{b}$ be two vectors such that $\left|\overline{a}\right|=2022,\left|\overline{b}\right|=2022and\overline{a}.\overline{b}=0$ for some $\alpha ,\beta \in R,\overline{c}=\alpha \overline{a}+\beta \overline{b}+\frac{\overline{a}\times \overline{b}}{2022}$ . If $\left|\overline{c}\right|=2022\sqrt{2}$ and the vector $\overline{c}$ makes an angle $\text{'}\theta \text{'}$  With both $\overline{a}and\overline{b},$ then $\left[2024{\cos }^2\theta \right]$ is….. (where [.] represents G.I.F)
 

$\overline{c}=\alpha \overline{a}+\beta \overline{b}+\frac{\overline{a}\times \overline{b}}{2022}\Rightarrow \overline{c}.\overline{a}=\alpha a^2\Rightarrow \alpha =\sqrt{2}\cos \theta and\overline{c}.\overline{b}=\beta b^2\Rightarrow \beta =\sqrt{2}\cos \theta$

${|\overline{a}\times \overline{b}|}^2={\left|\overline{a}\right|}^2{\left|\overline{b}\right|}^2-{(\overline{a}.\overline{b})}^2$

${|\overline{a}\times \overline{b}|}^2={\left(2022\right)}^4$

$\Rightarrow {|\overline{a}\times \overline{b}|}^2={\left(2022\right)}^2 andnow\overline{c}.\overline{c}={\alpha }^2{a}^2+{\beta }^2{b}^2+{\left(2022\right)}^2$

$\Rightarrow {\left(2022\right)}^2\left(2\right)={\left(2022\right)}^2.[2{\cos }^2\theta +2{\cos }^2\theta +1]$

$\Rightarrow 2=4{\cos }^2\theta +1\Rightarrow 4{\cos }^2\theta =1$

$\therefore \left[2024{\cos }^2\theta \right]=\left[506\left(4{\cos }^2\theta \right)\right]=506$