The null space of $A$ is the set of solutions to $A{\bf x}={\bf 0}$. To find this, you may take the augmented matrix $[A|0]$ and row reduce to an echelon form. Note that every entry in the rightmost column of this matrix will always be 0 in the row reduction steps. So, we may as well just row reduce $A$, and when finding solutions to $A{\bf x}={\bf 0}$, just keep in mind that the missing column is all 0's.
Suppose after doing this, you obtain
$$
\left[\matrix{1&0&0&0&-1 \cr 0&0&1&1&0 \cr 0&0&0&0&0 \cr 0&0&0&0&0 \cr }\right]
$$
Now, look at the columns that do not contain any of the leading row entries. These columns correspond to the free variables of the solution set to $A{\bf x}={\bf 0}$ Note that at this point, we know the dimension of the null space is 3, since there are three free variables. That the null space has dimension 3 (and thus the solution set to $A{\bf x}={\bf 0}$ has three free variables) could have also been obtained by knowing that the dimension of the column space is 2 from the rank-nullity theorem.
The "free columns" in question are 2,4, and 5. We may assign any value to their corresponding variable.
So, we set $x_2=a$, $x_4=b$, and $x_5=c$, where $a$, $b$, and $c$ are arbitrary.
Now solve for $x_1$ and $x_3$:
The second row tells us $x_3=-x_4=-b$ and the first row tells us $x_1=x_5=c$.
So, the general solution to $A{\bf x}={\bf 0}$ is
$$
{\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right]
$$
Let's pause for a second. We know:
1) The null space of $A$ consists of all vectors of the form $\bf x $ above.
2) The dimension of the null space is 3.
3) We need three independent vectors for our basis for the null space.
So what we can do is take $\bf x$ and split it up as follows:
$$\eqalign{
{\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right]
&=\left[ \matrix{0\cr a\cr 0\cr 0\cr 0}\right]+
\left[\matrix{c\cr 0\cr 0\cr 0\cr c}\right]+
\left[\matrix{0\cr 0\cr -b\cr b\cr 0}\right]\cr
&=
a\left[ \matrix{0\cr1\cr0\cr 0\cr 0}\right]+
c\left[ \matrix{1\cr 0\cr 0\cr 0\cr 1}\right]+
b\left[ \matrix{0\cr 0\cr -1\cr 1\cr 0}\right]\cr
}
$$
Each of the column vectors above are in the null space of $A$. Moreover, they are independent. Thus, they form a basis.
I'm not sure that this answers your question. I did a bit of "hand waving" here. What I glossed over were the facts:
1)The columns of the echelon form of $A$ that do not contain leading row entries correspond to the "free variables" to $A{\bf x}={\bf 0}$. If the number of these columns is $r$, then the dimension of the null space is $r$ (again, if you know the dimension of the column space, you can see that the dimension of the null space must be the number of these columns from the rank-nullity theorem).
2) If you split up the general solution to $A{\bf x}={\bf 0}$ as done above, then these vectors will be independent (and span of course since you'll have $r$ of them).
Notice that $v_{1} = v_{3} - v_{2}$. So $\{v_{1}, v_{4}\}$ is your basis, as you can form $v_{2}$ and $v_{3}$ from linear combinations of $v_{1}, v_{4}$.
Edit: I'll add a bit more on a basis. Note that a basis is a maximally independent set of vectors that spans the space. In $\mathbb{R}^{3}$, we would have three basis vectors. For subspaces of any vector space, we are permitted fewer vectors. A subspace is a vector space itself, so the number of basis vectors describes the dimension of the subspace.
Now in general vector spaces, we can use the determinant test to see if a set of vectors spans. The determinant test is nice, because (for a matrix $M$) $det(M) = 0$ if and only if the set of vectors in $M$ are linearly dependent. So if $det(M) \neq 0$ and $dim(M) = dim(V)$, for $V$ your vector space, then the vectors in $M$ form a basis.
In a subspace, you generally have fewer than $dim(V)$ vectors. I say generally, as a vector space is trivially a subspace of itself. So you can row-reduce a matrix to find the number of independent vectors. Or you can solve systems of linear equations yourself. If you are given vectors, these are your two options. If you are choosing your own basis without such constraints, selecting a subset of the standard basis is the way to go.
Best Answer
Intuitively: Since your basis vectors are linearly independent, they form a basis. The only problem is if they were spanning only a sub-lattice of the lattice you are describing and not the whole thing. This would be the case if you had chosen for example:
$$v_1'=2v_1=(6,-8,0,0,2)\qquad v_2=(2,-3,0,1,0)\qquad v_3=(1,-2,1,0,0)$$
Since there are obviously no other lattice points between 0 and the 3 points you have selected, everything is fine and the selected basis of $v_1,v_2,v_3$ spans the whole lattice.
Formally: Let $x=(x_1,x_2,x_3,x_4,x_5)\in L$ where L is your lattice. We need to prove that x can be written in the form $x=av_1+bv_2+cv_3=(3a+2b+c,-4a-3b-2c,c,b,a)$. Obviously, $x_3=c,x_4=b,x_5=a$. We also know that $$x_1+x_2=-x_3-x_4-x_5=-c-b-a$$ and $$x_1+2x_2=-3x_3-4x_4-5x_5=-3c-4b-5a$$ from the defining properties of the lattice, which can be solved to give $$x_1=3a+2b+c \quad\mbox{and}\quad x_2=-4a-3b-2c$$ as above. This way we have proven that any lattice point is of this form. Since we also knew that any vector of this form is a lattice point (because it satisfies the defining relations), we are done.