Step 1: Understanding the Question:
We need to find the normal unit vector for a plane defined by three points. There are two such vectors (opposite directions).
Step 2: Key Formula or Approach:
Unit normal $\hat{n} = \pm \frac{\vec{AB} \times \vec{AC}}{|\vec{AB} \times \vec{AC}|}$.
Step 3: Detailed Explanation:
Let $A = (1, -1, 2), B = (2, 0, -1), C = (0, 2, 1)$.
$\vec{AB} = (2-1)\hat{i} + (0-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + \hat{j} - 3\hat{k}$.
$\vec{AC} = (0-1)\hat{i} + (2-(-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$.
$\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -3\\ -1 & 3 & -1 \end{vmatrix}$
$\vec{n} = \hat{i}(-1 - (-9)) - \hat{j}(-1 - 3) + \hat{k}(3 - (-1))$
$\vec{n} = 8\hat{i} + 4\hat{j} + 4\hat{k}$.
Simplifying the direction (factor out 4): $\vec{d} = 2\hat{i} + \hat{j} + \hat{k}$.
Magnitude $|\vec{d}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}$.
The unit vectors are $\hat{n} = \pm \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$.
Step 4: Final Answer:
The unit vectors are $\pm \left( \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \right)$.