Example 6 Given that $a, b, c$ are three non-negative real numbers, and satisfy $3a+2b+c=5, 2a+b-3c=1$. If $S=$ $3a+b-7c$, then the sum of the maximum and minimum values of $S$ is $\qquad$
Analysis: There are three variables in $S$, and we should try to eliminate two variables through the method of elimination to facilitate finding the extremum.

Solution: From $3a + 2b = 5 - c$ and $2a + b = 3c + 1$, we get $a = 7c - 3, b = 7 - 11c$.
Thus, $S = 3(7c - 3) + 7 - 11c - 7c = 3c - 2$.
Since $a, b, c$ are non-negative real numbers, we have
$$
\left\{\begin{array}{l}
7c - 3 \geqslant 0, \\
7 - 11c \geqslant 0, \\
c \geqslant 0.
\end{array}\right.
$$

Solving these inequalities, we get $\frac{3}{7} \leqslant c \leqslant \frac{7}{11}$.
When $c = \frac{7}{11}$, $S$ has its maximum value $-\frac{1}{11}$; when $c = \frac{3}{7}$, $S$ has its minimum value $-\frac{5}{7}$.

Therefore, the sum of the maximum and minimum values of $S$ is $-\frac{62}{77}$.