3. $a, b, c$ are non-negative real numbers, and satisfy $3a+2b+c=$ $5, 2a+b-3c=1$. Let $m=3a+b-7c$, and let $x$ be the minimum value of $m$, $y$ be the maximum value of $m$. Then $xy=$ $\qquad$
$3 \cdot \frac{5}{77}$.
From $3a + 2b + c = 5, 2a + b - 3c = 1$, we get 
\[
\left\{
\begin{array}{l}
3a + 2b = 5 - c, \\
2a + b = 1 + 3c
\end{array}
\Rightarrow
\left\{
\begin{array}{l}
3a + 2b = 5 - c, \\
4a + 2b = 2 + 6c
\end{array}
\right.
\right.
\]
Thus, $a = 7c - 3, b = 7 - 11c$.
Since $a, b, c$ are non-negative real numbers, we have
\[
\left\{
\begin{array}{l}
7c - 3 \geqslant 0, \\
7 - 11c \geqslant 0, \\
c \geqslant 0
\end{array}
\Rightarrow \frac{3}{7} \leqslant c \leqslant \frac{7}{11}.
\right.
\]

Also, $m = 3a + b - 7c = 3c - 2$, so $-\frac{5}{7} \leqslant m \leqslant -\frac{1}{11}$.
Thus, $x = -\frac{5}{7}, y = -\frac{1}{11}$. Therefore, $xy = \frac{5}{77}$.