1. Given vectors

$\vec a=\hat i-3\hat j+7\hat k=(1,-3,7),\qquad \vec b=2\hat i-\hat j+\hat k=(2,-1,1)$

We are given

$(\vec a+2\vec b)\times \vec c=3(\vec c\times \vec a)$

and

$\vec a\cdot \vec c=130.$

We need to find $\vec b\cdot \vec c.$

2. Simplify the cross product equation

Use the anti-commutative property:

$\vec c\times \vec a=-(\vec a\times \vec c).$

So,

$(\vec a+2\vec b)\times \vec c=3(\vec c\times \vec a)=-3(\vec a\times \vec c).$

Bring everything to one side:

$(\vec a+2\vec b)\times \vec c+3\vec a\times \vec c=\vec 0.$

Using distributivity of cross product,

$\vec a\times \vec c+2\vec b\times \vec c+3\vec a\times \vec c=\vec 0$

$4\vec a\times \vec c+2\vec b\times \vec c=\vec 0$

Divide by $2$:

$2\vec a\times \vec c+\vec b\times \vec c=\vec 0.$

Now combine:

$(2\vec a+\vec b)\times \vec c=\vec 0.$

Hence, $2\vec a+\vec b$ is parallel to $\vec c$.

So we can write

$\vec c=\lambda(2\vec a+\vec b)$

for some scalar $\lambda$.

3. Compute $2\vec a+\vec b$

$2\vec a=2(1,-3,7)=(2,-6,14)$

Therefore,

$2\vec a+\vec b=(2,-6,14)+(2,-1,1)=(4,-7,15).$

Thus,

$\vec c=\lambda(4,-7,15).$

4. Use the condition $\vec a\cdot \vec c=130$

Since

$\vec a=(1,-3,7),\qquad \vec c=\lambda(4,-7,15),$

we get

$\vec a\cdot \vec c=\lambda\,(1,-3,7)\cdot(4,-7,15).$

Now,

$(1,-3,7)\cdot(4,-7,15)=1\cdot 4+(-3)(-7)+7\cdot 15$

$=4+21+105=130.$

So,

$\vec a\cdot \vec c=130\lambda=130.$

Hence,

$\lambda=1.$

Therefore,

$\vec c=(4,-7,15).$

5. Find $\vec b\cdot \vec c$

$\vec b=(2,-1,1),\qquad \vec c=(4,-7,15).$

So,

$\vec b\cdot \vec c=2\cdot 4+(-1)(-7)+1\cdot 15$

$=8+7+15=30.$

6. Final answer

$\boxed{30}$