To solve the problem, we need to calculate the determinant \( \Delta \) of the matrix formed by the three 3-digit even numbers represented as \( (x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3) \).
### Step-by-step Solution:
1. **Understanding the Matrix**:
The determinant \( \Delta \) is given by:
\[
\Delta = \begin{vmatrix}
x_1 & y_1 & z_1 \\
x_2 & y_2 & z_2 \\
x_3 & y_3 & z_3
\end{vmatrix}
\]
2. **Expanding the Determinant**:
We can expand this determinant using the formula:
\[
\Delta = x_1 \begin{vmatrix} y_2 & z_2 \\ y_3 & z_3 \end{vmatrix} - y_1 \begin{vmatrix} x_2 & z_2 \\ x_3 & z_3 \end{vmatrix} + z_1 \begin{vmatrix} x_2 & y_2 \\ x_3 & y_3 \end{vmatrix}
\]
This gives us:
\[
\Delta = x_1 (y_2 z_3 - y_3 z_2) - y_1 (x_2 z_3 - x_3 z_2) + z_1 (x_2 y_3 - x_3 y_2)
\]
3. **Identifying Even Numbers**:
Since \( x_1, y_1, z_1, x_2, y_2, z_2, x_3, y_3, z_3 \) are all 3-digit even numbers, the last digits \( z_1, z_2, z_3 \) must be even. Therefore, \( z_1, z_2, z_3 \) can be any of the even digits: 0, 2, 4, 6, 8.
4. **Properties of Even Numbers**:
- The product of two even numbers is even.
- The product of an even number and an odd number is even.
- The difference of two even numbers is even.
5. **Analyzing Each Term**:
- The term \( y_2 z_3 - y_3 z_2 \) is even since both \( y_2, y_3 \) and \( z_2, z_3 \) are even.
- The term \( x_2 z_3 - x_3 z_2 \) is also even for the same reason.
- The term \( x_2 y_3 - x_3 y_2 \) is even as well.
6. **Final Calculation**:
Since all terms in the determinant expansion are even, the entire expression for \( \Delta \) is even. Thus, we can conclude:
\[
\Delta \text{ is even}
\]
7. **Conclusion**:
Since \( \Delta \) is even, it is divisible by 2 but we cannot conclude that it is divisible by 4 or any higher power of 2 without additional information.
### Final Answer:
\[
\Delta \text{ is divisible by } 2 \text{ but not necessarily by } 4.
\]
