1. Given vectors

$\vec a=(3,1,-2),\quad \vec b=(4,1,7),\quad \vec c=(1,-3,4)$

We need a vector $\vec p$ such that

$\vec p\times \vec b=\vec c\times \vec b$

and

$\vec p\cdot \vec a=0.$

We must find

$\vec p\cdot (\hat i-\hat j-\hat k).$

2. Use the cross-product condition

From

$\vec p\times \vec b=\vec c\times \vec b,$

we get

$(\vec p-\vec c)\times \vec b=\vec 0.$

So $\vec p-\vec c$ is parallel to $\vec b$. Hence

$\vec p=\vec c+\lambda \vec b$

for some scalar $\lambda$.

Thus,

$\vec p=(1,-3,4)+\lambda(4,1,7)=(1+4\lambda,-3+\lambda,4+7\lambda).$

3. Use the dot-product condition

Given

$\vec p\cdot \vec a=0,$

so

$(\vec c+\lambda \vec b)\cdot \vec a=0.$

First compute:

$\vec c\cdot \vec a=(1)(3)+(-3)(1)+(4)(-2)=3-3-8=-8.$

Next,

$\vec b\cdot \vec a=(4)(3)+(1)(1)+(7)(-2)=12+1-14=-1.$

Therefore,

$-8+\lambda(-1)=0$

$-8-\lambda=0$

$\lambda=-8.$

So,

$\vec p=\vec c-8\vec b.$

Now calculate $\vec p$:

$\vec p=(1,-3,4)-8(4,1,7)=(1-32,-3-8,4-56)=(-31,-11,-52).$

4. Find the required dot product

We need

$\vec p\cdot (1,-1,-1).$

So,

$(-31,-11,-52)\cdot (1,-1,-1)=-31+11+52=32.$

5. Check options

The value is

$\boxed{32}$

So the correct option is B.