Let $ x$, $ y$, $ z$ be integers such that
\[ \begin{array}{l} {x \minus{} 3y \plus{} 2z \equal{} 1} \\
{2x \plus{} y \minus{} 5z \equal{} 7} \end{array}
\]
Then $ z$ can be

$\textbf{(A)}\  3^{111} \qquad\textbf{(B)}\  4^{111} \qquad\textbf{(C)}\  5^{111} \qquad\textbf{(D)}\  6^{111} \qquad\textbf{(E)}\ \text{None}$
To solve the given system of linear equations for \(x\), \(y\), and \(z\), we start by writing the equations in matrix form and then use the method of elimination or substitution.

The given system of equations is:
\[
\begin{array}{l}
x - 3y + 2z = 1 \\
2x + y - 5z = 7
\end{array}
\]

1. **Express one variable in terms of the others:**
   From the first equation, solve for \(x\):
   \[
   x = 1 + 3y - 2z
   \]

2. **Substitute \(x\) into the second equation:**
   Substitute \(x = 1 + 3y - 2z\) into the second equation:
   \[
   2(1 + 3y - 2z) + y - 5z = 7
   \]
   Simplify:
   \[
   2 + 6y - 4z + y - 5z = 7
   \]
   Combine like terms:
   \[
   7y - 9z = 5
   \]

3. **Solve for \(y\) in terms of \(z\):**
   \[
   y = \frac{5 + 9z}{7}
   \]
   For \(y\) to be an integer, \(\frac{5 + 9z}{7}\) must be an integer. This implies that \(5 + 9z\) must be divisible by 7.

4. **Find \(z\) modulo 7:**
   We need to find \(z\) such that \(5 + 9z \equiv 0 \pmod{7}\):
   \[
   9z \equiv -5 \pmod{7}
   \]
   Simplify \(9 \pmod{7}\):
   \[
   2z \equiv -5 \pmod{7}
   \]
   Since \(-5 \equiv 2 \pmod{7}\):
   \[
   2z \equiv 2 \pmod{7}
   \]
   Divide both sides by 2:
   \[
   z \equiv 1 \pmod{7}
   \]

5. **General solution for \(z\):**
   The general solution for \(z\) is:
   \[
   z = 7n + 1 \quad \text{for some integer } n
   \]

6. **Check the given options:**
   We need to check which of the given options satisfies \(z \equiv 1 \pmod{7}\):
   \[
   3^{111} \pmod{7}, \quad 4^{111} \pmod{7}, \quad 5^{111} \pmod{7}, \quad 6^{111} \pmod{7}
   \]

   Using Fermat's Little Theorem, \(a^{p-1} \equiv 1 \pmod{p}\) for a prime \(p\):
   \[
   3^6 \equiv 1 \pmod{7}, \quad 4^3 \equiv 1 \pmod{7}, \quad 5^6 \equiv 1 \pmod{7}, \quad 6^6 \equiv 1 \pmod{7}
   \]

   Calculate each:
   \[
   3^{111} \equiv 3^{111 \mod 6} \equiv 3^3 \equiv 27 \equiv 6 \pmod{7}
   \]
   \[
   4^{111} \equiv 4^{111 \mod 3} \equiv 4^0 \equiv 1 \pmod{7}
   \]
   \[
   5^{111} \equiv 5^{111 \mod 6} \equiv 5^3 \equiv 125 \equiv 6 \pmod{7}
   \]
   \[
   6^{111} \equiv 6^{111 \mod 6} \equiv 6^3 \equiv 216 \equiv 6 \pmod{7}
   \]

   Only \(4^{111} \equiv 1 \pmod{7}\).

The final answer is \(\boxed{4^{111}}\)