To find the area of the triangle with vertices A(1, -1, 2), B(2, 1, -1), and C(3, -1, 2), we can use the formula for the area of a triangle given by three points in 3D space. The area can be calculated using the cross product of two vectors formed by these points.
### Step-by-Step Solution:
1. **Define the Points as Vectors:**
Let the points A, B, and C be represented as vectors:
\[
\vec{A} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}, \quad \vec{B} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}, \quad \vec{C} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}
\]
2. **Find the Vectors AB and AC:**
We can find the vectors AB and AC:
\[
\vec{AB} = \vec{B} - \vec{A} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}
\]
\[
\vec{AC} = \vec{C} - \vec{A} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}
\]
3. **Calculate the Cross Product AB × AC:**
The area of the triangle is given by half the magnitude of the cross product of vectors AB and AC. We calculate the cross product:
\[
\vec{AB} \times \vec{AC} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} \times \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}
\]
Using the determinant method:
\[
\vec{AB} \times \vec{AC} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -3 \\
2 & 0 & 0
\end{vmatrix}
\]
4. **Calculate the Determinant:**
Expanding the determinant:
\[
= \hat{i} \begin{vmatrix} 2 & -3 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -3 \\ 2 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix}
\]
\[
= \hat{i}(2 \cdot 0 - (-3) \cdot 0) - \hat{j}(1 \cdot 0 - (-3) \cdot 2) + \hat{k}(1 \cdot 0 - 2 \cdot 2)
\]
\[
= 0\hat{i} + 6\hat{j} - 4\hat{k}
\]
So,
\[
\vec{AB} \times \vec{AC} = \begin{pmatrix} 0 \\ 6 \\ -4 \end{pmatrix}
\]
5. **Find the Magnitude of the Cross Product:**
The magnitude of the cross product is:
\[
|\vec{AB} \times \vec{AC}| = \sqrt{0^2 + 6^2 + (-4)^2} = \sqrt{0 + 36 + 16} = \sqrt{52} = 2\sqrt{13}
\]
6. **Calculate the Area of the Triangle:**
The area of triangle ABC is given by:
\[
\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} (2\sqrt{13}) = \sqrt{13}
\]
### Final Answer:
The area of the triangle with vertices A(1, -1, 2), B(2, 1, -1), and C(3, -1, 2) is \(\sqrt{13}\) square units.
