To find the area of a triangle given the three vertices, $A = (0, 1, -2)$, $B = (2, 1, 1)$, and $C = (-2, 1, 1)$, we can use the determinant formula for the area of a triangle in 3D space. The area $A$ is calculated using:

$\text{Area} = \frac{1}{2} \times |\vec{AB} \times \vec{AC}|$

Step 1: Calculate vectors $\vec{AB}$ and $\vec{AC}$.

$\vec{AB} = B - A = (2 - 0, 1 - 1, 1 - (-2)) = (2, 0, 3)$ $\vec{AC} = C - A = (-2 - 0, 1 - 1, 1 - (-2)) = (-2, 0, 3)$

Step 2: Compute the cross product $\vec{AB} \times \vec{AC}$.

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & 3 \\ -2 & 0 & 3 \end{vmatrix}$

This results in:

$\vec{AB} \times \vec{AC} = (0 \cdot 3 - 3 \cdot 0) \mathbf{i} - (2 \cdot 3 - 3 \cdot (-2)) \mathbf{j} + (2 \cdot 0 - 0 \cdot (-2)) \mathbf{k}$

Simplifying:

$\vec{AB} \times \vec{AC} = (0) \mathbf{i} - (6 + 6) \mathbf{j} + (0) \mathbf{k} = -12 \mathbf{j}$

Step 3: Find the magnitude of the cross product.

$|\vec{AB} \times \vec{AC}| = \sqrt{0^2 + (-12)^2 + 0^2} = \sqrt{144} = 12$

Step 4: Calculate the area of the triangle.

$\text{Area} = \frac{1}{2} \times 12 = 6$

Therefore, the area of the triangle is 6 square units.

Now, looking at the answer choices, we can see that none of the provided options exactly match this calculation. However, the closest matching form from the options seems to be option a: $4\sqrt{6}$.

Do you want to check any details or ask for more clarification? Here are five related questions:

1. How is the determinant method applied in 3D geometry?
2. What are some properties of the cross product in 3D space?
3. How can vector geometry be used to calculate distances and areas?
4. Why is the magnitude of the cross product proportional to the area of a triangle?
5. How does the formula for a triangle's area in 2D differ from the 3D case?

Tip: The cross product of two vectors gives a new vector that is perpendicular to both original vectors and whose magnitude represents the area of the parallelogram formed by the two vectors.