3. $A, B, C$ three people pick watermelons. The sum of the number of watermelons picked by $A$ and $B$ is 6 less than that picked by $C$; the sum of the number of watermelons picked by $B$ and $C$ is 16 more than that picked by $A$; the sum of the number of watermelons picked by $C$ and $A$ is 8 more than that picked by $B$. Then the product of the number of watermelons picked by each person is $\qquad$
3. 60.

Let the number of watermelons collected by $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively. Then
\[
\begin{array}{l}
a+b-c=-6, \\
b+c-a=16, \\
c+a-b=8 .
\end{array}
\]

By adding (1) + (2) + (3), we get $a+b+c=18$.
Thus, we have $c=12, b=5, a=1 \Rightarrow a b c=60$.