3. In a rectangular coordinate system, there is $\triangle A B C$, with vertices $A$ $(3,0)$, $B(0,2)$, and $C(0,-1)$. The area of the common part between $\triangle A B C$ and the two parallel lines $x=t$, $x=\frac{t}{2}(0<t \leqslant 3)$ is denoted as $S(t)$. Then, as $t$ varies, the maximum value of $S(t)$ is . $\qquad$
3. $\frac{3}{2}$.

Notice that
$$
\begin{array}{l}
l_{A B}: \frac{x}{3}+\frac{y}{2}=1(0 \leqslant x \leqslant 3), \\
l_{A C}: \frac{x}{3}-y=1(0 \leqslant x \leqslant 3) .
\end{array}
$$

As shown in Figure 3, let the intersection points of the line $x=t$ with segments $A B$ and $A C$ be $F$ and $E$ respectively; the intersection points of the line $x=\frac{t}{2}$ with segments $A B$ and $A C$ be $G$ and $H$ respectively. Then
$$
\begin{array}{c}
y_{F}=2-\frac{2 t}{3}, \\
y_{E}=\frac{t}{3}-1, y_{G}=2-\frac{t}{3}, y_{H}=\frac{t}{6}-1 .
\end{array}
$$

Notice that
$$
\begin{array}{c}
S(t)=\frac{1}{2} \times \frac{t}{2}\left\{\left[\left(2-\frac{2 t}{3}\right)-\left(\frac{t}{3}-1\right)\right]+\right. \\
\left.\left[\left(2-\frac{t}{3}\right)-\left(\frac{t}{6}-1\right)\right]\right\} \\
=-\frac{3}{8}\left(t^{2}-4 t\right)=-\frac{3}{8}(t-2)^{2}+\frac{3}{2} .
\end{array}
$$

Therefore, when $t=2$, $S(t)_{\max }=\frac{3}{2}$.