1. Given vectors

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

We need to find $\vec c \cdot \vec b$ given:

$\vec b \times \vec c = \vec b \times \vec a$

and

$\vec c \cdot \vec a = 0$

2. Use the cross product condition

From

$\vec b \times \vec c = \vec b \times \vec a$

we get

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

This means $\vec c-\vec a$ is parallel to $\vec b$. Hence,

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

for some scalar $\lambda$.

3. Use the dot product condition

Given

$\vec c \cdot \vec a = 0$

Substitute $\vec c = \vec a + \lambda \vec b$:

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

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

Now compute:

$\vec a\cdot \vec a = 1^2+(-2)^2+1^2 = 1+4+1=6$

$\vec b\cdot \vec a = 1\cdot 1 + (-1)(-2) + 1\cdot 1 = 1+2+1=4$

So,

$6+4\lambda=0$

$\lambda = -\frac{6}{4} = -\frac{3}{2}$

Thus,

$\vec c = \vec a - \frac{3}{2}\vec b$

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

$\vec c\cdot \vec b = \left(\vec a - \frac{3}{2}\vec b\right)\cdot \vec b$

$= \vec a\cdot \vec b - \frac{3}{2}(\vec b\cdot \vec b)$

We already have:

$\vec a\cdot \vec b = 4$

Now,

$\vec b\cdot \vec b = 1^2+(-1)^2+1^2 = 1+1+1=3$

Therefore,

$\vec c\cdot \vec b = 4 - \frac{3}{2}\cdot 3 = 4 - \frac{9}{2} = \frac{8-9}{2} = -\frac{1}{2}$

5. Check options

$\vec c\cdot \vec b = -\frac{1}{2}$

So the correct option is A.