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