Example 6 Given three non-negative real numbers $a, b, c$ satisfying the conditions: $3a+2b+c=5, 2a+b-3c=1$. Let $s=$ $3a+b-7c$ have a maximum value of $M$ and a minimum value of $m$. Find $\mathrm{Mm}$.
(1989, Shanghai Junior High School Mathematics Competition)
Solution: From the given conditions, we have $a=7c-3, b=7-$ $11c$, substituting into $s=3a+b-7c$, we get $s=3c-2$.
Since $a, b, c$ are non-negative real numbers,
$$
\therefore\left\{\begin{array}{l}
a = 7c - 3 \geqslant 0, \\
b = 7 - 11c \geqslant 0
\end{array} \Rightarrow \left\{\begin{array}{l}
c \geqslant \frac{3}{7}, \\
c \leqslant \frac{7}{11}.
\end{array}\right.\right.
$$

That is, $\frac{3}{7} \leqslant c \leqslant \frac{7}{11}$.
Thus, $-\frac{5}{7} \leqslant s \leqslant -\frac{1}{11}$.

Solving, we get $M=-\frac{1}{11}, m=-\frac{5}{7}$.
Therefore, $M m=\frac{5}{77}$.