Q. 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
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