To determine which system of equations has the solution [tex]\((-2, 0, 1)\)[/tex], we'll test each system by substituting [tex]\(x = -2\)[/tex], [tex]\(y = 0\)[/tex], and [tex]\(z = 1\)[/tex] into each equation and checking if all the equations in the system are satisfied.
### System A
[tex]\[
\begin{aligned}
x + y & = -2 \\
-2 + 0 & = -2 \quad \text{(True)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
y - z & = 1 \\
0 - 1 & = -1 \quad \text{(False)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
x + y - z & = -1 \\
-2 + 0 - 1 & = -3 \quad \text{(False)}
\end{aligned}
\][/tex]
Since not all equations are satisfied, System A does not have [tex]\((-2, 0, 1)\)[/tex] as a solution.
### System B
[tex]\[
\begin{aligned}
x + y & = -2 \\
-2 + 0 & = -2 \quad \text{(True)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
y - z & = 1 \\
0 - 1 & = -1 \quad \text{(False)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
z + y - z & = -1 \\
1 + 0 - 1 & = y = 0 \quad \text{(False, this simplifies to } y = 0 \text{, not a valid equation)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
3x - y & = -6 \\
3(-2) - 0 & = -6 \quad \text{(True)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
2y - 5z & = -5 \\
2(0) - 5(1) & = -5 \quad \text{(True)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
z - y - 2z & = -4 \\
1 - 0 - 2(1) & = -1 \quad \text{(False)}
\end{aligned}
\][/tex]
Since not all equations are satisfied, System B does not have [tex]\((-2, 0, 1)\)[/tex] as a solution.
### System C
[tex]\[
\begin{aligned}
-2x + 2y & = 4 \\
-2(-2) + 2(0) & = 4 \quad \text{(True)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
y + 7z & = -7 \\
0 + 7(1) & = 7 \quad \text{(False)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
x - y - z & = -3 \\
-2 - 0 - 1 & = -3 \quad \text{(True)}
\end{aligned}
\][/tex]
Since not all equations are satisfied, System C does not have [tex]\((-2, 0, 1)\)[/tex] as a solution.
### System D
[tex]\[
\begin{aligned}
-3x + 2y & = 6 \\
-3(-2) + 2(0) & = 6 \quad \text{(True)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
y + 2z & = 2 \\
0 + 2(1) & = 2 \quad \text{(True)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
z - y - z & = -1 \\
1 - 0 - 1 & = 0 \quad \text{(False)}
\end{aligned}
\][/tex]
Since not all equations are satisfied, System D does not have [tex]\((-2, 0, 1)\)[/tex] as a solution.
### Conclusion
None of the systems (A, B, C, or D) satisfy the solution [tex]\((-2, 0, 1)\)[/tex] fully. Therefore, the answer is that none of the given systems of equations have [tex]\((-2, 0, 1)\)[/tex] as a solution.
