5. Given that $x, y, z$ are non-negative real numbers, and satisfy
$$
\left\{\begin{array}{l}
3 x+2 y+z=5 . \\
2 x+y-3 z=1 .
\end{array}\right.
$$

Let $s=3 x+y-7 z$. Then the maximum value of $s$ is ( ).
(A) $-\frac{1}{11}$
(B) $\frac{1}{11}$
(C) $-\frac{5}{7}$
(D) $-\frac{7}{5}$
5. A.

From the problem, we have
$$
\left\{\begin{array}{l}
x=7 z-3 \geqslant 0, \\
y=7-11 z \geqslant 0
\end{array} \Rightarrow \frac{3}{7} \leqslant z \leqslant \frac{7}{11}\right. \text {. }
$$

Then $s=3 x+y-7 z$
$$
\begin{array}{l}
=3(7 z-3)+(7-11 z)-7 z \\
=3 z-2 .
\end{array}
$$

Therefore, $s_{\max }=3 \times \frac{7}{11}-2=-\frac{1}{11}$.