5. When three non-negative real numbers $x, y, z$ satisfy the equations $x+3 y+2 z=3$ and $3 x+3 y+z=4$, the minimum and maximum values of $M=3 x-2 y+4 z$ are ( ).
(A) $-\frac{1}{7}, 6$
(B) $-\frac{1}{6}, 7$
(C) $\frac{1}{5}, 8$
(D) $-\frac{1}{8}, 5$
5. B.

From $\left\{\begin{array}{l}x+3 y+2 z=3, \\ 3 x+3 y+z=4,\end{array}\right.$ we get
$$
\left\{\begin{array}{l}
y=\frac{5}{3}(1-x), \\
z=2 x-1 .
\end{array}\right.
$$

Substituting into the expression for $M$ yields
$$
\begin{aligned}
M=3 & -2 y+4 z \\
& =3 x-\frac{10}{3}(1-x)+4(2 x-1) \\
& =\frac{43}{3} x-\frac{22}{3} .
\end{aligned}
$$

Since $x$, $y$, and $z$ are all non-negative real numbers, we have
$$
\left\{\begin{array}{l}
x \geqslant 0, \\
\frac{5}{3}(1-x) \geqslant 0, \\
2 x-1 \geqslant 0,
\end{array}\right.
$$

which means $\frac{1}{2} \leqslant x \leqslant 1$.
Therefore, when $x=\frac{1}{2}$, $M_{\text {min }}=-\frac{1}{6}$;
when $x=1$, $M_{\text {max }}=7$.