Number of vectors of unit length perpendicular to $\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{j} + \hat{k}$ is:

Vector perpendicular to both $\vec{a}$ and $\vec{b}$ is along $\vec{a} \times \vec{b}$. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1 \end{vmatrix} = (\hat{i})(1 - 2) - (\hat{j})(2 - 0) + (\hat{k})(2 - 0) = -\hat{i} - 2\hat{j} + 2\hat{k}$ Unit vector in this direction can be $\pm \dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$ → There are **two** such unit vectors. $\boxed{\text{Answer: (B) two}}$