1. We need a unit vector $\hat{c}$ such that:

   • $\hat{c}$ lies in the plane of $\vec a$ and $\vec b$
   • $\hat{c} \perp \vec a$

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

2. Since $\hat{c}$ lies in the plane of $\vec a$ and $\vec b$, let $\hat{c} = \alpha \vec a + \beta \vec b$ for some scalars $\alpha, \beta$.

3. Because $\hat{c}$ is perpendicular to $\vec a$, $\hat{c} \cdot \vec a = 0$ $(\alpha \vec a + \beta \vec b) \cdot \vec a = 0$ $\alpha (\vec a \cdot \vec a) + \beta (\vec b \cdot \vec a) = 0$

4. Compute the dot products: $\vec a \cdot \vec a = 1^2 + 2^2 + 1^2 = 6$ $\vec b \cdot \vec a = 2\cdot 1 + 1\cdot 2 + (-1)\cdot 1 = 2+2-1=3$

   So, $6\alpha + 3\beta = 0$ $2\alpha + \beta = 0 \implies \beta = -2\alpha$

5. Therefore, $\hat{c} = \alpha \vec a - 2\alpha \vec b = \alpha(\vec a - 2\vec b)$

   Now, $\vec a - 2\vec b = (1,2,1) - 2(2,1,-1) = (1,2,1) - (4,2,-2) = (-3,0,3)$

   Hence the required direction is $(-3,0,3) = 3(-1,0,1)$ so a unit vector is

   = \frac{1}{\sqrt2}(-\hat i + \hat k)$$

6. Check with options:

   • A: $\frac{1}{\sqrt2}(-\hat i + \hat k)$ ✓
   • B, C, D: do not satisfy the required conditions.

Therefore, the correct answer is: $\boxed{\frac{1}{\sqrt2}(-\hat i + \hat k)}$