Let $\overline{a}=\widehat{i}+\widehat{j}+\widehat{k},$ $\overline{b}=-\widehat{i}-8\widehat{j}+2\widehat{k}$   and  $\overline{c}=4\widehat{i}+c_2\widehat{j}+c_3\widehat{k}$  be three vectors such that  $\overline{b}\times \overline{a}=\overline{c}\times \overline{a.}$ If the angle between the vector  $\overline{c}$  and the vector  $3\widehat{i}+4\widehat{j}+\widehat{k}$  is  $\theta ,$ then the  greatest integer less than or equal to  ${\tan }^2\theta$  is_____

Given  $\overline{b}\times \overline{a}=\overline{c}\times \overline{a}$
 $(\overline{b}-\overline{c})\times \overline{a}=0$
$\therefore \overline{b}-\overline{c}$   is parallel to  $\overline{a}$
 $$\begin{gathered} \overline{b}-\overline{c}=\lambda \overline{a} \\ \overline{c}=\overline{b}-\lambda \overline{a} \\ \overline{c}=(-1-\lambda )\widehat{i}+(-8-\lambda )\widehat{j}+(2-\lambda )\widehat{k} \end{gathered}$$
Given  $\overline{c}=4\widehat{i}+c_2\widehat{j}+c_3\widehat{k}$
$$\begin{gathered} 4=-1-\lambda \\ \lambda =5 \\ \therefore \overline{c}=4\widehat{i}-3\widehat{j}+7\widehat{k} \\ \cos \theta =\frac{\overline{c}(3\widehat{i}+4\widehat{j}+\widehat{k})}{\overline{\left|c\right|}\sqrt{3^2+4^2+1^2}} \\ =\frac{12-12+7}{\sqrt{74}\sqrt{26}} \\ =\frac{7}{2\sqrt{481}} \end{gathered}$$ 
Given angle between  $\overline{c}$  &   $3\widehat{i}+4\widehat{j}+\widehat{k}$   $\theta$
 $$\begin{gathered} \therefore \tan \theta =\frac{\sqrt{1875}}{7} \\ {\tan }^2\theta =\frac{1879}{49} \\ ~38.2 \\ =38 \end{gathered}$$