Let  $\overrightarrow{a}=2\widehat{i}-\widehat{j}+2\widehat{k}$  and  $\overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k}$ . Let a vector $\overrightarrow{v}$ be in the plane containing  $\overrightarrow{a}$  and $\overrightarrow{b}$ . If  $\overrightarrow{v}$ is perpendicular to the vector $3\widehat{i}+2\widehat{j}-\widehat{k}$  and its projection on  $\overrightarrow{a}$  is 19 units, then $|2\overrightarrow{v}{|}^2$   is equal to_________.

$$\begin{gathered} \left(1494\right):Let\overrightarrow{v}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}=\lambda (2\widehat{i}-\widehat{j}+2\widehat{k})+\mu (\widehat{i}+2\widehat{j}-\widehat{k})\Rightarrow \overrightarrow{v}=(2\lambda +\mu )\widehat{i}+(2\mu -\lambda )\widehat{j}+(2\lambda -\mu )\widehat{k} \\ Now,\overrightarrow{v}.(3\widehat{i}+2\widehat{j}-\widehat{k}=0) \Rightarrow 3(2\lambda +\mu )+2(2\mu -\lambda )+\mu -2\lambda =0 \Rightarrow 2\lambda +8\mu =0\Rightarrow \lambda =-4\mu We\text{\hspace{0.17em}}have,\frac{\overrightarrow{v}.\overrightarrow{a}}{\overrightarrow{a}}=19 \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{[(2\lambda +\mu )\widehat{i}+(2\mu -\lambda )\widehat{j}+(2\lambda -\mu )\widehat{k}].(2\widehat{i}-\widehat{j}+2\widehat{k})}{3}=19 \\ \Rightarrow 57=4\lambda +2\mu +\lambda -2\mu +4\lambda -2\mu \\ \Rightarrow 9\mu -2\mu =57\Rightarrow -38\mu =57\Rightarrow \mu =\frac{-3}{2}and\lambda =6 \\ \therefore \overrightarrow{v}=(12-\frac{3}{2})\widehat{i}-9\widehat{j}+(12+\frac{3}{2})\widehat{k}\Rightarrow 2\overrightarrow{v}=21\widehat{i}-18\widehat{j}+27\widehat{k} \\ \therefore {\left|2\overrightarrow{v}\right|}^2={21}^2+{(-18)}^2+{27}^2=1494 \end{gathered}$$