Let $\stackrel{\wedge }{a},\stackrel{\wedge }{b}$ be unit vectors . If $\overrightarrow{\mathrm{c}}$be a vector such that the angle between $\widehat{\mathrm{a}} \mathrm{and} \overrightarrow{\mathrm{c}} \mathrm{is} \frac{\mathrm{\pi }}{12}\mathrm{, }$and $\widehat{\mathrm{b}}=\overrightarrow{\mathrm{c}}+2(\overrightarrow{\mathrm{c}}\times \widehat{\mathrm{a}}),\text{ then }|6\overrightarrow{\mathrm{c}}{|}^2$is equal to

$\begin{array}{l}|\widehat{\mathrm{b}}{|}^2=|\overrightarrow{\mathrm{c}}+2(\overrightarrow{\mathrm{c}}\times \widehat{\mathrm{a}}){|}^2\\ |\widehat{\mathrm{b}}{|}^2=|\mathrm{c}{|}^2+4|\overrightarrow{\mathrm{c}}\times \widehat{\mathrm{a}}{|}^2+4\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{c}}\times \widehat{\mathrm{a}})\\ \Rightarrow 1=|\mathrm{c}{|}^2+4|\mathrm{c}{|}^2{\sin }^2\frac{\mathrm{\pi }}{12}+0\\ \Rightarrow 1=|\mathrm{c}{|}^2+4|\mathrm{c}{|}^2{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}^2\\ \therefore 1=|\mathrm{c}{|}^2\left(1+\frac{4-2\sqrt{3}}{2}\right)\Rightarrow 1=\left(3-\sqrt{3}\right)|\mathrm{c}{|}^2\\ \Rightarrow |\mathrm{c}{|}^2=\frac{1}{3-\sqrt{3}}=\frac{3+\sqrt{3}}{6}\end{array}$
So $6^2|\mathrm{c}{|}^2=6(3+\sqrt{3})$