1. Given data

We have | a|=\sqrt{31}, \qquad 4|\vec b|=2, \qquad |\vec c|=2. So, $|\vec b|=\frac12, \qquad |\vec c|=2.$

Also, $2(\vec a\times \vec b)=3(\vec c\times \vec a).$ Since $\vec c\times \vec a=-(\vec a\times \vec c),$ this becomes $2(\vec a\times \vec b)=-3(\vec a\times \vec c).$ Hence, $\vec a\times (2\vec b+3\vec c)=0.$

Therefore, $2\vec b+3\vec c$ is parallel to $\vec a$. So there exists some scalar $\lambda$ such that $\vec a=\lambda(2\vec b+3\vec c).$

2. Use magnitude of $\vec a$ to find $\lambda$

First compute $|2\vec b+3\vec c|^2=4|\vec b|^2+9|\vec c|^2+12(\vec b\cdot \vec c).$

Now, $|\vec b|=\frac12, \quad |\vec c|=2, \quad \angle(\vec b,\vec c)=\frac{2\pi}{3},$ so $\vec b\cdot \vec c=|\vec b||\vec c|\cos\frac{2\pi}{3}=\frac12\cdot 2\cdot\left(-\frac12\right)=-\frac12.$

Thus, $|2\vec b+3\vec c|^2=4\left(\frac14\right)+9(4)+12\left(-\frac12\right)=1+36-6=31.$ So, $|2\vec b+3\vec c|=\sqrt{31}.$

But $|\vec a|=\sqrt{31}$ too, and $\vec a=\lambda(2\vec b+3\vec c).$ Hence, $|\lambda|=1.$ So we may take $\vec a=\pm(2\vec b+3\vec c).$ The required quantity is squared, so sign will not matter.

3. Compute $\vec a\cdot \vec b$

Using $\vec a=2\vec b+3\vec c$, $\vec a\cdot \vec b=(2\vec b+3\vec c)\cdot \vec b=2|\vec b|^2+3(\vec c\cdot \vec b).$ Now,

and $3(\vec b\cdot \vec c)=3\left(-\frac12\right)=-\frac32.$ Therefore, $\vec a\cdot \vec b=\frac12-\frac32=-1.$ If $\vec a=-(2\vec b+3\vec c)$, then this becomes $1$; in either case, the square is same.

So, $(\vec a\cdot \vec b)^2=1.$

4. Compute $|\vec a\times \vec c|$

Again using $\vec a=2\vec b+3\vec c$, $\vec a\times \vec c=(2\vec b+3\vec c)\times \vec c=2(\vec b\times \vec c)+3(\vec c\times \vec c)=2(\vec b\times \vec c).$ Thus,

Now, $|\vec b\times \vec c|=|\vec b||\vec c|\sin\frac{2\pi}{3}=\frac12\cdot 2\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{2}.$ Hence, $|\vec a\times \vec c|=2\cdot \frac{\sqrt3}{2}=\sqrt3.$ So, $|\vec a\times \vec c|^2=3.$

5. Required value

The expression means

Therefore, $\frac{3}{1}=3.$

Final Answer

$\boxed{3}$