Show that the $\mathbb{Z}$-module $A=\{(x,y,z)\in\mathbb{Z}^3: 2x+3y+3z=0\}$ is free and find a base.
I tried to show that the set $\{(3,-2,0),(3,0,-2)\}$ generates $A$ but there seems to be a problem since I find that $$(x,y,z)=-\dfrac{y}{2}(3,-2,0)-\dfrac{z}{2}(3,0,-2)$$
From $2x+3y+3z=0 \Rightarrow 2|y+z$ but can I have $2|y$ and $2|z$?
In the general case
When is the $\mathbb{Z}$-module $B=\{(x,y,z)\in\mathbb{Z}^3: ax+by+cz=0,a,b,c\in\mathbb{Z}\}$ free and how can I find a base for it?
We claim that if $S = \{2x+3y+3z = 0\}$ and $T = (3m,j,-2m-j), m,j \in \mathbb Z$, then $S = T$.
Note that $2(3m) + 3(j) + 3(-2m-j) = 6m + 3j -6m-3j = \implies T \subset S$.
Conversely, if $2x+3y+3z = 0$, then $3 | x$ and $2 | y+z$. Now, write $j = y$, $x = 3m$ and $z = -2m-j$, where $m,j$ are integers to get that $S \subset T$.
Now, $T = (3m,j,-2m-j) = m(3,0,-2) + j(0,1,-1)$ for $m,j$ arbitrary integers, so $T$ is a free module , therefore so is $S$.
Try to see how the second part of the argument above worked out, to work out how the solutions of $ax+by+cz=0$ form a free module. Note that you can assume that WLOG $a,b,c$ don't share any common factors.