3. Given

point set $T=\{(x, y) \mid | x+1|+| y-2 | \leqslant 3\}$,
number set $M=\{x+2 y \mid(x, y) \in T\}$.
Then the sum of the maximum and minimum elements in set $M$ is . $\qquad$
3. 6 .

The point set $T$ is the quadrilateral $A B C D$ and its interior points formed by four lines
$$
\begin{array}{l}
(x+1)+(y-2)=3, (x+1)-(y-2)=3, \\
-(x+1)+(y-2)=3, \\
-(x+1)-(y-2)=3
\end{array}
$$

where $A(-1,5), B(-4,2), C(-1,-1), D(2,2)$.

Let $x+2 y=b$, i.e., $y=-\frac{1}{2} x+\frac{1}{2} b$. From the graph, it can be seen that this line passes through points $A$ and $C$ when $\frac{1}{2} b$ takes its maximum and minimum values, respectively.
Therefore, when $x=-1, y=5$,
$b_{\text {max }}=x+2 y=9$;
when $x=-1, y=-1$,
$b_{\text {min }}=x+2 y=-3$.
Thus, the required sum is 6.