Let's represent in the Cartesian coordinate system the set of points \((x, y)\) defined by the equation

$$
3|x-1|+2|y+2|=6
$$
(1) On the left side, neither of the two non-negative terms can be greater than 6. Thus, $x$ and $y$ can take values within the following bounds (see figure):

$$
-1 \leq x \leq 3 \quad -5 \leq y \leq 1
$$

The points satisfying the conditions (2) lie inside the rectangle whose side equations are:

$$
x=-1, \quad x=3, \quad y=-5, \quad y=1
$$

In equation (1), we can replace $|x-1|$ with $+(x-1)$ if $x-1>0$; if $x-1<0$, then $|x-1|=-(x-1)$. A similar observation can be made for $|y+2|$. Therefore, from (1), the following 4 line equations are derived:

I.

II.

III.

IV.

$$
\begin{aligned}
3(x-1)+2(y+2) & =6 \\
-3(x-1)+2(y+2) & =6 \\
3(x-1)-2(y+2) & =6 \\
-3(x-1)-2(y+2) & =6
\end{aligned}
$$

Thus, the geometric place we are looking for consists of the points on the segments of the lines given by (3) that lie inside or on the boundary of the rectangle described by (2).