Find all linear integer relations of following vectors I have following excercise:

a) Find all linear integer relations of $(7,1,0),(0,3,-1),(-2,-2,3),(1,-1,-1)$.
b) Let $M$ be the submodule generated by these vectors in $\mathbb{Z^3}$. Is $\mathbb{Z^3/}M$ free? Is it finitely generated?

I have had a couple of questions like this during my course and I am never able to do them. Sometimes it is requiered to find relations of integers and sometimes, like in this case, of these vectors. The theory we are doing is very vague so I really don't know where to take my information.
About a) I thought about writing something like $x(7,1,0)+y(0,3,-1)+z(-2,-2,3)+m(1,-1,-1)=0$
Then I get:
$7x-2z+m=0$
$x+3y-2z-m=0$
$-y+3z-m=0$
And then I can get a matrix, which I don't think I can reduce in RREF since I can only do integer operations. So I got completely stuck.
About b) I am clueless. I get panicked everyime I see a quotient module and have to determine if it is free or not so I would be immensely great if somebody could help me out.
Thanks in advance.
 A: I can't help with part (a), but I think I have a solution for part (b). Let's call the vectors you mentioned $v_1,v_2,v_3$ and $v_4$. Assuming by "created" you mean "generated by", i.e. $M = \langle v_1,v_2,v_3,v_4\rangle$, we see the following:
$$
v_3 + 2v_4 = \begin{pmatrix}0\\-4\\1\end{pmatrix} \implies \begin{pmatrix}1\\-4\\1\end{pmatrix}+v_2 = \begin{pmatrix}0\\-1\\0\end{pmatrix} \implies (-1) \cdot\begin{pmatrix}0\\-1\\0\end{pmatrix} = \begin{pmatrix}0\\1\\0\end{pmatrix} =: e_2.
$$
Similarly, we get
$$
v_2 - 3e_2 = \begin{pmatrix}0\\0\\-1\end{pmatrix} \implies (-1) \cdot\begin{pmatrix}0\\0\\-1\end{pmatrix} = \begin{pmatrix}0\\0\\1\end{pmatrix} =: e_3.
$$
Lastly, we get
$$
v_1 - 6v_4 = \begin{pmatrix}1\\7\\6\end{pmatrix} \implies \begin{pmatrix}1\\7\\6\end{pmatrix} - 7e_2 - 6e_3 = \begin{pmatrix}1\\0\\0\end{pmatrix} =: e_1.
$$
Hence, we see that $e_1,e_2,e_3 \in M$, so  $M = \mathbb{Z}^3$, since $B := \{e_1,e_2,e_3\}$ is a base for $\mathbb{Z}^3$ as a $\mathbb{Z}$-module. Therefore, we simply get $\mathbb{Z}^3/M = \{0\}$, the trivial module. $\{0\}$ is obviously free. It is finitely generated by the basis $\emptyset$. Please correct me if I got anything wrong!
