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