Let A B C be a triangle such that B C → = a → , C A → = b → , A B → = c &#8594…

We know that by triangle law of vector addition,

$\overrightarrow{a}+\overrightarrow{\mathit{b}}+\overrightarrow{\mathit{c}}=0$

$\Rightarrow \overrightarrow{b}+\overrightarrow{c}=-\overrightarrow{a}$

Squaring both side we get,

${\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2+2\overrightarrow{b}·\overrightarrow{c}={\left|\overrightarrow{a}\right|}^2$

Now putting the given value $\left|\overrightarrow{a}\right|=6\sqrt{2}, \left|\overrightarrow{b}\right|=2\sqrt{3} \& \overrightarrow{b}.\overrightarrow{c}=12$

We get, ${\left|\overrightarrow{c}\right|}^2=36$

So, $\left|\overrightarrow{c}\right|=6$

Now, $S1: \left|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{c}\times \overrightarrow{b}\right|-\left|\overrightarrow{c}\right|$

$\Rightarrow \left|\left(\overrightarrow{a}+\overrightarrow{c}\right)\times \overrightarrow{b}\right|-\left|\overrightarrow{c}\right|$

$\Rightarrow 0-6=-6$

And $S2: \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$

$\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$

Squaring both side we get,

${\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\angle ACB\right)={\left|\overrightarrow{c}\right|}^2$

$\Rightarrow 72+12-24\sqrt{6}\cos \left(\angle ACB\right)=36$

$\Rightarrow \cos \left(\angle ACB\right)=\frac{48}{24\sqrt{6}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}$