To find the area of triangle ABC with vertices A(1, 1, 1), B(3, 7, 4), and C(-1, 3, 0), we can use the formula for the area of a triangle formed by three points in three-dimensional space. The area can be calculated using the cross product of two vectors formed by the points.
### Step-by-Step Solution:
1. **Define the Points**:
Let the points be:
- \( A(1, 1, 1) \)
- \( B(3, 7, 4) \)
- \( C(-1, 3, 0) \)
2. **Find the Vectors**:
We need to find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):
- \( \overrightarrow{AB} = B - A = (3 - 1, 7 - 1, 4 - 1) = (2, 6, 3) \)
- \( \overrightarrow{AC} = C - A = (-1 - 1, 3 - 1, 0 - 1) = (-2, 2, -1) \)
3. **Calculate the Cross Product**:
The area of triangle ABC is given by:
\[
\text{Area} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right|
\]
To find the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \), we set up the determinant:
\[
\overrightarrow{AB} \times \overrightarrow{AC} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 3 \\
-2 & 2 & -1
\end{vmatrix}
\]
Calculating this determinant:
\[
= \hat{i} \begin{vmatrix} 6 & 3 \\ 2 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 3 \\ -2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 6 \\ -2 & 2 \end{vmatrix}
\]
\[
= \hat{i} (6 \cdot -1 - 3 \cdot 2) - \hat{j} (2 \cdot -1 - 3 \cdot -2) + \hat{k} (2 \cdot 2 - 6 \cdot -2)
\]
\[
= \hat{i} (-6 - 6) - \hat{j} (-2 + 6) + \hat{k} (4 + 12)
\]
\[
= -12 \hat{i} - 4 \hat{j} + 16 \hat{k}
\]
4. **Magnitude of the Cross Product**:
Now, we find the magnitude of the vector:
\[
\left| \overrightarrow{AB} \times \overrightarrow{AC} \right| = \sqrt{(-12)^2 + (-4)^2 + (16)^2}
\]
\[
= \sqrt{144 + 16 + 256} = \sqrt{416}
\]
5. **Calculate the Area**:
Finally, we can find the area of triangle ABC:
\[
\text{Area} = \frac{1}{2} \sqrt{416} = \frac{\sqrt{416}}{2}
\]
Simplifying \( \sqrt{416} \):
\[
\sqrt{416} = \sqrt{16 \times 26} = 4\sqrt{26}
\]
Thus, the area becomes:
\[
\text{Area} = \frac{4\sqrt{26}}{2} = 2\sqrt{26}
\]
### Final Answer:
The area of triangle ABC is \( 2\sqrt{26} \) square units.
