Let ABC be a triangle such that BC = a, CA = b, AB = c,| a |=6 √ 2,| b |=2 √ 3 and b · c =12. Consider the statements: (S1): |( a × b )+( c × b )|-|…

Let ABC be a triangle such that $\displaystyle \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},|\overrightarrow{\mathrm{a}}|=6 \sqrt{2},|\overrightarrow{\mathrm{~b}}|=2 \sqrt{3}$ and $\displaystyle \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=12$. Consider the statements : (S1) : $\displaystyle |(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})|-|\overrightarrow{\mathrm{c}}|=6(2 \sqrt{2}-1)$ (S2) : $\displaystyle \angle \mathrm{ACB}=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$ Then