The correct answer is $\boxed{\text{(A)}}$.
To solve the system of equations, we can use the elimination method. First, we add the first two equations together. This gives us $2x + 2y + 2z = 4$. Then, we subtract the third equation from this equation. This gives us $-x = -1$. Solving for $x$, we get $x = 1$.
Now that we know $x = 1$, we can substitute this value into any of the original equations to solve for $y$ and $z$. Substituting into the first equation, we get $1 + y + z = 4$. Solving for $y$, we get $y = 2$. Substituting into the second equation, we get $1 – 2 + z = 0$. Solving for $z$, we get $z = 1$.
Therefore, the solution of the system of equations is $\boxed{x = 1, y = 2, z = 1}$.
Here is a brief explanation of each option:
- Option (A): $x = 2, y = 2, z = 0$. This is the correct answer.
- Option (B): $x = 1, y = 4, z = 1$. This is not the correct answer. To see why, we can substitute $x = 1$, $y = 4$, and $z = 1$ into the first equation. This gives us $1 + 4 + 1 = 6$, which is not equal to 4. Therefore, this option is not the correct answer.
- Option (C): $x = 2, y = 4, z = 3$. This is not the correct answer. To see why, we can substitute $x = 2$, $y = 4$, and $z = 3$ into the second equation. This gives us $2 – 4 + 3 = 1$, which is not equal to 0. Therefore, this option is not the correct answer.
- Option (D): $x = 1, y = 2, z = 1$. This is the correct answer.