Let ABC be a triangle such that $\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},\left|\overrightarrow{\mathrm{a}}\right|=6\sqrt{2},\left|\overrightarrow{\mathrm{b}}\right|=2\sqrt{3}$ and $\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}}=12$

Consider the statements :

$\left(\mathrm{S}1\right):\left|\right(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{b}}\left)\right|-\left|\overrightarrow{\mathrm{c}}\right|=6(2\sqrt{2}-1)$

$\left(\mathrm{S}2\right):\mathrm{\angle }\mathrm{ABC}={\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right).$ Then 

$\begin{array}{l}\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0\\ \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}}\\ |\overrightarrow{\mathrm{b}}{|}^2+|\overrightarrow{\mathrm{c}}{|}^2+2\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{a}}{|}^2\\ |\overrightarrow{\mathrm{c}}{|}^2=36\\ \left|\overrightarrow{\mathrm{c}}\right|=6\\ \mathrm{S}1:|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{b}}|-\left|\overrightarrow{\mathrm{c}}\right|\\ \left|\right(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}})\times \overrightarrow{\mathrm{b}}|-\left|\overrightarrow{\mathrm{c}}\right|\\ |-\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{b}}|-\left|\overrightarrow{\mathrm{c}}\right|\\ 0-\left|\overrightarrow{\mathrm{c}}\right|=-6\Rightarrow \left|\overrightarrow{\mathrm{c}}\right|=6\\ \mathrm{S}2:\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0\\ \\ |\overrightarrow{\mathrm{a}}{|}^2+|\overrightarrow{\mathrm{b}}{|}^2-2\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\cos \left(\mathrm{\angle }\mathrm{ACB}\right)=|\overrightarrow{\mathrm{c}}{|}^2\\ \cos \left(\mathrm{\angle }\mathrm{ACB}\right)=\sqrt{\frac{2}{3}}\\ \overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}}=\left|\overline{b}\right|\left|\overline{c}\right|\cos \angle ABC=12\\ \cos \angle ABC=\frac{12}{\left(2\sqrt{3}\right)6}=\frac{1}{\sqrt{3}}\end{array}$