For the matrix:
$$ \begin{bmatrix} 1 & 0 & 2 & | & 0 \\ 0 & 1 & 3 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} $$
which means $x_1 + 0x_2 + 2x_3 = 0$ and $x_2 + 3x_3 = 0$.
It has $3$ eqns, and $3$ variables are to be determined. This means $x_2$ is free to vary. So I get the solution : $[-2t, t, -t/3]$ for $t$ any real number.
But how do I determine the basis? How could this be described graphically?
The solution is actually of the form $(\frac{2}{3}t, t, -\frac{1}{3}t)$. So one possible basis can be $(\frac{2}{3}, 1, -\frac{1}{3})$ or $(2, 3, -1)$ if you like integers...