Let $$\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$$ and $$\vec{b} = \hat{i} - \hat{j} + \hat{k}$$ be two vectors. If $$\vec{c}$$ is a vector such that $$\vec{b} \times \vec{c} = \vec{b} \times \vec{a}$$ and $$\vec{c} \cdot \vec{a} = 0$$, then $$\vec{c} \cdot \vec{b}$$ is equal to.

We have the two given vectors

$$\vec a = \hat i - 2\hat j + \hat k \qquad\text{and}\qquad \vec b = \hat i - \hat j + \hat k.$$

A third vector $$\vec c$$ satisfies the two simultaneous conditions

$$\vec b \times \vec c = \vec b \times \vec a \qquad\text{and}\qquad \vec c \cdot \vec a = 0.$$

First, we employ the basic property of the cross product: if two vectors have the same cross product with a non-zero vector $$\vec b,$$ then their difference must be parallel to $$\vec b.$$ Stating the property explicitly:

Formula: $$\vec p \times \vec q = \vec 0 \; \Longrightarrow \; \vec p \parallel \vec q.$$\ Applying it, we write

$$\vec b \times \left(\vec c - \vec a\right) = \vec 0 \;\; \Longrightarrow \;\; \vec c - \vec a \parallel \vec b.$$

So we can express $$\vec c$$ as

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

where $$\lambda$$ is some real scalar that we still need to determine.

Now we impose the second condition $$\vec c \cdot \vec a = 0.$$ Substituting $$\vec c = \vec a + \lambda\vec b$$ gives

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

Expanding the dot product:

$$\vec a\cdot\vec a + \lambda\,\vec b\cdot\vec a = 0.$$

We calculate the two scalar products individually. First, using $$\vec a = (1,\,-2,\,1),$$ we have

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

Next, using $$\vec b = (1,\,-1,\,1),$$ we obtain

$$\vec b\cdot\vec a = 1\cdot1 + (-1)(-2) + 1\cdot1 = 1 + 2 + 1 = 4.$$

Substituting these numerical values into the relation $$\vec a\cdot\vec a + \lambda\,\vec b\cdot\vec a = 0$$ gives

$$6 + \lambda\,(4) = 0.$$

So,

$$4\lambda = -6 \;\;\Longrightarrow\;\; \lambda = -\frac32.$$

Having found $$\lambda,$$ we now require $$\vec c\cdot\vec b.$$ First write $$\vec c$$ explicitly:

$$\vec c = \vec a + \lambda\,\vec b = (1, -2, 1) + \left(-\frac32\right)(1, -1, 1).$$

However, to avoid component-wise arithmetic twice, it is quicker to compute the dot product directly using the linearity of the dot product:

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

We already know $$\vec a\cdot\vec b = 4.$$ Next, find $$\vec b\cdot\vec b$$ using $$\vec b = (1,\,-1,\,1):$$

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

Therefore,

$$\vec c\cdot\vec b = 4 + \lambda\,(3).$$

Substitute $$\lambda = -\dfrac32:$$

$$\vec c\cdot\vec b = 4 + \left(-\frac32\right)(3) = 4 - \frac92 = \frac{8}{2} - \frac{9}{2} = -\frac12.$$

Thus the required scalar product is

$$\vec c\cdot\vec b = -\frac12.$$

Hence, the correct answer is Option C.