Let $$ABC$$ be a triangle such that $$\vec{BC} = \vec{a}$$, $$\vec{CA} = \vec{b}$$, $$\vec{AB} = \vec{c}$$, $$|\vec{a}| = 6\sqrt{2}$$, $$|\vec{b}| = 2\sqrt{3}$$ and $$\vec{b} \cdot \vec{c} = 12$$. Consider the statements:
$$S_1: |\vec{a} \times (\vec{b} + \vec{c})| \times |\vec{b} - \vec{c}| = 6(2\sqrt{2} - 1)$$
$$S_2: \angle ABC = \cos^{-1}\sqrt{\dfrac{2}{3}}$$
Then

Given,

$$\vec{BC}=\vec a,\qquad \vec{CA}=\vec b,\qquad \vec{AB}=\vec c$$

Also,

$$|\vec a|=6\sqrt2,\qquad |\vec b|=2\sqrt3,\qquad \vec b\cdot\vec c=12$$

Since

$$\vec a+\vec b+\vec c=0,$$

we get

$$\vec c=-(\vec a+\vec b)$$

Now,

$$\vec b\cdot\vec c=12$$

$$\vec b\cdot(-\vec a-\vec b)=12$$

$$-\vec a\cdot\vec b-|\vec b|^2=12$$

$$-\vec a\cdot\vec b-(2\sqrt3)^2=12$$

$$-\vec a\cdot\vec b-12=12$$

$$\vec a\cdot\vec b=-24$$

Now find

$$|\vec c|^2$$

$$|\vec c|^2=|\vec a+\vec b|^2$$

$$=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b$$

$$=(6\sqrt2)^2+(2\sqrt3)^2+2(-24)$$

$$=72+12-48$$

$$=36$$

Hence,

$$|\vec c|=6$$

Now check statement $$S_1.$$

Since

$$\vec b+\vec c=-\vec a,$$

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

Therefore,

$$|\vec a\times(\vec b+\vec c)|=0$$

Hence,

$$|\vec a\times(\vec b+\vec c)|\times|\vec b-\vec c|=0$$

which is not equal to

$$6(2\sqrt2-1)$$

Therefore,

$$S_1$$ is false.

Now check $$S_2.$$

Angle $$ABC$$ is the angle between

$$\vec{BA}=-\vec c$$

and

$$\vec{BC}=\vec a$$

Hence,

$$\cos\angle ABC=\frac{(-\vec c)\cdot\vec a}{|\vec c||\vec a|}$$

Now,

$$\vec c=-(\vec a+\vec b)$$

$$\vec a\cdot\vec c=\vec a\cdot(-\vec a-\vec b)$$

$$=-|\vec a|^2-\vec a\cdot\vec b$$

$$=-72-(-24)$$

$$=-48$$

Thus,

$$(-\vec c)\cdot\vec a=48$$

Therefore,

$$\cos\angle ABC=\frac{48}{6\cdot6\sqrt2}$$

$$=\frac{48}{36\sqrt2}$$

$$=\frac4{3\sqrt2}$$

$$=\frac{2\sqrt2}{3}$$

$$=\sqrt{\frac89}$$

which is not equal to

$$\sqrt{\frac23}$$

Hence,

$$S_2$$ is false.

Therefore,

$$\boxed{\text{Both }S_1\text{ and }S_2\text{ are false}}$$.