Let ABC be a triangle such that → BC = → a , → CA = → b , → AB = →...

We know that by triangle law of vector addition,

$\overset{\to}{a} + \overset{\to}{b} + \overset{\to}{c} = 0$

$⇒ \overset{\to}{b} + \overset{\to}{c} = - \overset{\to}{a}$

Squaring both side we get,

${\left|\overset{\to}{b}\right|}^{2} + {\left|\overset{\to}{c}\right|}^{2} + 2\overset{\to}{b} \cdot \overset{\to}{c} = {\left|\overset{\to}{a}\right|}^{2}$

Now putting the given value \left|\overset{\to}{a}\right| = 6\sqrt{2}, \left|\overset{\to}{b}\right| = 2\sqrt{3} \text{ and } \overset{\to}{b} . \overset{\to}{c} = 12$</p><p>We get,${\left|\overset{\to}{c}\right|}^{2} = 36$</p><p>So, \left|\overset{\to}{c}\right| = 6$

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

$⇒ \left|\left(\overset{\to}{a} + \overset{\to}{c}\right) \times \overset{\to}{b}\right| - \left|\overset{\to}{c}\right|$

$⇒ 0 - 6 = - 6$

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

$\overset{\to}{a} + \overset{\to}{b} = - \overset{\to}{c}$

Squaring both side we get,

${\left|\overset{\to}{a}\right|}^{2} + {\left|\overset{\to}{b}\right|}^{2} - 2\left|\overset{\to}{a}\right|\left|\overset{\to}{b}\right|\cos \left(∠ ACB\right) = {\left|\overset{\to}{c}\right|}^{2}$

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

$⇒ \cos \left(∠ ACB\right) = \dfrac{48}{24\sqrt{6}} = \dfrac{2}{\sqrt{6}} = \sqrt{\dfrac{2}{3}}$