Use the triangle vector relation

For triangle $ABC$, $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec 0$ so $\vec c+\vec a+\vec b=\vec 0 \quad \Rightarrow \quad \vec c=-(\vec a+\vec b).$

Given data

$|\vec a|=6\sqrt 2,\qquad |\vec b|=2\sqrt 3,\qquad \vec b\cdot \vec c=12.$

Since $\vec c=-(\vec a+\vec b)$, $\vec b\cdot \vec c=\vec b\cdot[-(\vec a+\vec b)] =-(\vec a\cdot\vec b)-|\vec b|^2.$

Therefore, $12=-(\vec a\cdot\vec b)-(2\sqrt3)^2=-(\vec a\cdot\vec b)-12$ which gives $\vec a\cdot\vec b=-24.$

Find the angle between $\vec a$ and $\vec b$

$\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ $-24=(6\sqrt2)(2\sqrt3)\cos\theta=12\sqrt6\cos\theta$ so $\cos\theta=-\frac{24}{12\sqrt6}=-\frac{2}{\sqrt6}=-\sqrt{\frac23}.$

Check statement (S2)

Angle $\angle ACB$ is the angle between $\overrightarrow{CA}=\vec b$ and $\overrightarrow{CB}=-\vec a$.

Hence, $\cos\angle ACB=\frac{\vec b\cdot(-\vec a)}{|\vec b||\vec a|}=-\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$ $=-\frac{-24}{12\sqrt6}=\frac{2}{\sqrt6}=\sqrt{\frac23}.$

Therefore, $\angle ACB=\cos^{-1}\left(\sqrt{\frac23}\right).$

So (S2) is true.

Find $|\vec c|$

Since $\vec c=-(\vec a+\vec b)$, $|\vec c|^2=|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b.$

Thus, $|\vec c|^2=(6\sqrt2)^2+(2\sqrt3)^2+2(-24)=72+12-48=36,$ so $|\vec c|=6.$

Check statement (S1)

Compute: $(\vec a\times \vec b)+(\vec c\times \vec b)=(\vec a+\vec c)\times \vec b.$

But from $\vec a+\vec b+\vec c=0$, $\vec a+\vec c=-\vec b.$

Hence, $(\vec a\times \vec b)+(\vec c\times \vec b)=(-\vec b)\times \vec b=\vec 0.$

Therefore, $\left| (\vec a\times \vec b)+(\vec c\times \vec b)\right|-|\vec c|=0-6=-6.$

But the RHS given in (S1) is $6(2\sqrt2-1)=12\sqrt2-6 \neq -6.$

So (S1) is false.

Conclusion

• (S1) is false
• (S2) is true

Therefore the correct option is: $\boxed{\text{C: only (S2) is true}}$