To preface, this is a study problem not a homework problem.
Let $A$ be a $7 \times 3$ matrix such that its null space is spanned by $[1,2,0]^T$, $[2,1,0]^T$, and $[1,-1,0]^T$.
The rank of A is: $1, 2, 3, 4, 6$?
So I know the answer is $1$ because, I've looked at the solution. I got the answer $2$.
What I did is the following: I found a basis for the null space by forming a matrix from the $3$ column vectors that span the null-space of $A$. The basis had dimension 1, so then nullity of $A$ is $1$. I know the $\operatorname{rank}(A) + \operatorname{nullity}(a)$ is the number of columns, so $3-1 = \operatorname{rank} = 2$
Where did I go wrong?
Notice that $(1,2,0)^T$ is not a multiple of $(2,1,0)^T$, the nullity is at least $2$.
Also, given that the third coordinate is $0$, we know that the null space has dimension $2$.
Hence, the rank is $3-2=1$.
Your mistake is claiming that the nullity is $1$.