To find the area of the triangle formed by the points $P = (0, 2, 1)$, $Q = (2, 0, 1)$, and $R = (3, -1, 2)$, we can use the following formula:

$\text{Area} = \frac{1}{2} \sqrt{|\mathbf{AB} \times \mathbf{AC}|^2}$

where $\mathbf{A} = P$, $\mathbf{B} = Q$, and $\mathbf{C} = R$.

First, find the vectors $\mathbf{AB}$ and $\mathbf{AC}$:

$\mathbf{AB} = Q - P = (2 - 0, 0 - 2, 1 - 1) = (2, -2, 0)$ $\mathbf{AC} = R - P = (3 - 0, -1 - 2, 2 - 1) = (3, -3, 1)$

Next, compute the cross product $\mathbf{AB} \times \mathbf{AC}$:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 0 \\ 3 & -3 & 1 \\ \end{vmatrix}$$ This expands to: $$\mathbf{AB} \times \mathbf{AC} = \mathbf{i}((-2)(1) - (0)(-3)) - \mathbf{j}((2)(1) - (0)(3)) + \mathbf{k}((2)(-3) - (-2)(3))$$ $$\mathbf{AB} \times \mathbf{AC} = \mathbf{i}(-2) - \mathbf{j}(2) + \mathbf{k}(-6 + 6)$$ $$\mathbf{AB} \times \mathbf{AC} = -2\mathbf{i} - 2\mathbf{j} + 0\mathbf{k}$$ $$\mathbf{AB} \times \mathbf{AC} = (-2, -2, 0)$$ Now, find the magnitude of the cross product: $$|\mathbf{AB} \times \mathbf{AC}| = \sqrt{(-2)^2 + (-2)^2 + 0^2} = \sqrt{4 + 4 + 0} = \sqrt{8} = 2\sqrt{2}$$ Finally, the area of the triangle is: $$\text{Area} = \frac{1}{2} \times 2\sqrt{2} = \sqrt{2}$$ So, the area of the triangle is $$\sqrt{2}$$ square units. Would you like more details or have any questions? Here are five related questions: 1. How is the cross product of two vectors determined? 2. What is the significance of the magnitude of the cross product in geometry? 3. How do you determine if three points are collinear using the area of a triangle? 4. Can the area formula be generalized to higher dimensions? 5. What happens to the area if all points lie in a 2D plane instead of 3D? **Tip:** The cross product of two vectors gives a vector that is perpendicular to the plane containing the original vectors, and its magnitude is proportional to the area of the parallelogram formed by the vectors.