1. **State the problem:** Find a unit vector perpendicular to both vectors $\mathbf{a} = 2\mathbf{i} + \mathbf{j} + \mathbf{k}$ and $\mathbf{b} = -2\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}$.\n\n2. **Formula used:** The vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ is given by their cross product $\mathbf{a} \times \mathbf{b}$.\n\n3. **Calculate the cross product:**\n$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ -2 & 3 & 2 \end{vmatrix} = \mathbf{i}(1 \cdot 2 - 1 \cdot 3) - \mathbf{j}(2 \cdot 2 - 1 \cdot (-2)) + \mathbf{k}(2 \cdot 3 - 1 \cdot (-2))$$\n$$= \mathbf{i}(2 - 3) - \mathbf{j}(4 + 2) + \mathbf{k}(6 + 2) = -\mathbf{i} - 6\mathbf{j} + 8\mathbf{k}$$\n\n4. **Find the magnitude of the cross product:**\n$$|\mathbf{a} \times \mathbf{b}| = \sqrt{(-1)^2 + (-6)^2 + 8^2} = \sqrt{1 + 36 + 64} = \sqrt{101}$$\n\n5. **Find the unit vector:** Divide the cross product by its magnitude.\n$$\mathbf{u} = \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} = \frac{-\mathbf{i} - 6\mathbf{j} + 8\mathbf{k}}{\sqrt{101}} = -\frac{1}{\sqrt{101}}\mathbf{i} - \frac{6}{\sqrt{101}}\mathbf{j} + \frac{8}{\sqrt{101}}\mathbf{k}$$\n\n**Final answer:** The unit vector perpendicular to both given vectors is\n$$\boxed{\mathbf{u} = -\frac{1}{\sqrt{101}}\mathbf{i} - \frac{6}{\sqrt{101}}\mathbf{j} + \frac{8}{\sqrt{101}}\mathbf{k}}$$