Let $\overline{a},\overline{b}$ be unit vectors .If $\overline{c}$ be a vector such that angle between $\overline{a}and\overline{c}$ is $\frac{\pi }{12}\text{\hspace{0.17em}\hspace{0.17em}}$ and  $\overline{b}=\overline{c}+2(\overline{c}\times \overline{a})$ then  ${\left|6\overline{c}\right|}^2=$

Given $\left|\overline{a}\right|=1,\left|\overline{b}\right|=1,(\overline{a},\overline{c})=\frac{\pi }{12}$

$\overline{b}=\overline{c}+2\left(\overline{c}\times \overline{a}\right)$

$\Rightarrow {\left|\overline{b}\right|}^2={\left|\overline{c}+2\left(\overline{c}\times \overline{a}\right)\right|}^2$

$\Rightarrow {\left|\overline{b}\right|}^2={\left|\overline{c}\right|}^2+4{\left|\overline{c}\times \overline{a}\right|}^2+4\left\{\overline{c}.\left(\overline{c}\times \overline{a}\right)\right\}$

$\Rightarrow {\left|\overline{b}\right|}^2={\left|\overline{c}\right|}^2+4{\left|\overline{c}\times \overline{a}\right|}^2+0$

$\Rightarrow {\left|\overline{b}\right|}^2={\left|\overline{c}\right|}^2+4{\left|\overline{c}\right|}^2{\left|\overline{a}\right|}^2{\sin }^2\left(\overline{c},\overline{a}\right)$

$\Rightarrow 1={\left|\overline{c}\right|}^2+4{\left|\overline{c}\right|}^2{\sin }^2\frac{\pi }{12}$

$\Rightarrow 1={\left|\overline{c}\right|}^2+4{\left|\overline{c}\right|}^2{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}^2$

$\Rightarrow 1={\left|\overline{c}\right|}^2+4{\left|\overline{c}\right|}^2\left(\frac{4-2\sqrt{3}}{8}\right)$

$\Rightarrow 1={\left|\overline{c}\right|}^2+4{\left|\overline{c}\right|}^2\left(\frac{2-\sqrt{3}}{4}\right)$

$\Rightarrow 4=4{\left|\overline{c}\right|}^2+4{\left|\overline{c}\right|}^2\left(2-\sqrt{3}\right)$

$\Rightarrow 4=4{\left|\overline{c}\right|}^2\left(1+2-\sqrt{3}\right)$

$\Rightarrow 1={\left|\overline{c}\right|}^2\left(3-\sqrt{3}\right)$

$\Rightarrow {\left|\overline{c}\right|}^2=\frac{1}{3-\sqrt{3}}\times \frac{3+\sqrt{3}}{3+\sqrt{3}}=\frac{3+\sqrt{3}}{6}$

$\Rightarrow 6{\left|\overline{c}\right|}^2=3+\sqrt{3}$

$\Rightarrow 6^2{\left|\overline{c}\right|}^2=6\left(3+\sqrt{3}\right)$

$\Rightarrow {\left|6\overline{c}\right|}^2=6\left(3+\sqrt{3}\right)$