Find a matrix whose null space is equal to the range of the matrix A? So here is the solution:

I don't understand how the null space relates to consistency of the matrix A. In another word, I don't understand how this method of solving leads to answer. 
 A: Notice that the range of the matrix $A$ is the span of the two column vectors:
$$\operatorname{span}(1,1,0)^T,(2,0,1)^T)$$
Now $(x,y,z)^T$ will be in the range of $A$ if the augmented matrix is not invertible so by the row echeloned matrix we see that this will be if
$$  -x+y+2z=0\iff [-1\; 1\;2](x,y,z)^T=0\iff(x,y,z)^T\in\ker [-1\; 1\;2]$$
A: This approach makes more sense to me since I can visualize it geometrically: 
The range of $A$ is the span of its two column vectors $(1, 1, 0)^T$ and $(2, 0, 1)^T$, which is a plane $P$ that passes through the origin in $\mathbf{R}^3$. We're then looking for a matrix $B$ whose nullspace is equal to this plane, i.e. $Bx = 0$ for any $x \in P$. It should be apparent that the $B$ that accomplishes this is $B = n^T$ where $n$ is the normal vector to the plane ($n$ is orthogonal to all points on the plane and thus the dot product of any point on the plane with $n$ is 0). You can compute $n$ by taking the cross product of the columns of $A$ which gives you $(2, 0, 1)^T \times (1, 1, 0)^T = (-1, 1, 2)^T$. Thus $B = (-1, 1, 2)$. Hope this makes sense.
