1. Given vectors

$\vec a = \hat i + 2\hat j - \hat k = (1,2,-1), \quad \vec b = \hat i - \hat j = (1,-1,0), \quad \vec c = \hat i - \hat j - \hat k = (1,-1,-1).$

We need to find $\vec r \cdot \vec a$ given:

$\vec r \times \vec a = \vec c \times \vec a$ and $\vec r \cdot \vec b = 0.$

2. Use the cross product condition

Given $\vec r \times \vec a = \vec c \times \vec a,$ we get $(\vec r - \vec c) \times \vec a = \vec 0.$

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

$\vec r = \vec c + \lambda \vec a$ for some scalar $\lambda$.

3. Apply the dot product condition

Now, $\vec r \cdot \vec b = 0.$ Substitute $\vec r = \vec c + \lambda \vec a$:

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

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

Compute these dot products:

$\vec c \cdot \vec b = (1,-1,-1) \cdot (1,-1,0) = 1 + 1 + 0 = 2,$

$\vec a \cdot \vec b = (1,2,-1) \cdot (1,-1,0) = 1 - 2 + 0 = -1.$

Thus, $2 + \lambda(-1) = 0 \implies 2 - \lambda = 0 \implies \lambda = 2.$

Therefore, $\vec r = \vec c + 2\vec a.$

4. Find $\vec r \cdot \vec a$

$\vec r \cdot \vec a = (\vec c + 2\vec a) \cdot \vec a = \vec c \cdot \vec a + 2(\vec a \cdot \vec a).$

Now,

$\vec c \cdot \vec a = (1,-1,-1) \cdot (1,2,-1) = 1 - 2 + 1 = 0,$

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

Hence, $\vec r \cdot \vec a = 0 + 2\times 6 = 12.$

5. Final answer

$\boxed{12}$