Are the span of these vectors the same? Let's say we have to solve the equation $2x+3y+z=0$. 
If we let $x=s, y=t,$ then $z=-2s-3t$, and the solution set is \begin{bmatrix}s\\t\\-2s-3t\end{bmatrix} which is the span of the vectors \begin{bmatrix}1\\0\\-2\end{bmatrix} and \begin{bmatrix}0\\1\\-3\end{bmatrix} However, if instead we had let  $y=t$ and $z=s$, we would have gotten 
\begin{bmatrix}-1.5s-.5t\\t\\s\end{bmatrix} or the span of the vectors 
\begin{bmatrix}-1\\0\\2\end{bmatrix} and \begin{bmatrix}-3\\2\\0\end{bmatrix}Are these results equivalent? 
 A: Yes they are. Notice that a plane has infinitely many vectors in it. So as long as we don't choose two vectors that are parallel to each other or none of them is $\vec{0}$, those two vectors will span the plane from which they are chosen (Since plane is 2-dimensional, we are chosing $2$ vectors). In other words, you can find infinitely many vector pairs on the same plane, which span that plane.
For example, in cartesian coordinate system,
$$\bigg\{\begin{bmatrix}1\\0\end{bmatrix},{\begin{bmatrix}0\\1\end{bmatrix}}\bigg\}$$
is the most common base. But
$$\bigg\{\begin{bmatrix}1\\1\end{bmatrix},{\begin{bmatrix}2\\1\end{bmatrix}}\bigg\}$$
also spans the same plane as well as
$$\bigg\{\begin{bmatrix}1\\0\end{bmatrix},{\begin{bmatrix}4\\1\end{bmatrix}}\bigg\}$$
and so on. 
In your case, notice that you can also combine the results that you have already found in order to have a different base as
$$\begin{bmatrix}0\\1\\-3\end{bmatrix}, \begin{bmatrix}-3\\2\\0\end{bmatrix}$$
which also spans the given plane.
A: The second pair of vectors is obviously linearly independent so also spans a two-dimensional space. Therefore, it suffices to show that both are in the span of the first pair of vectors. This should be obvious for $[-1,0,2]^T$. For the second vector of the two, you need a linear combination of $[1,0,-2]^T$ and $[0,1,-3]^T$ that annihilates the last coordinate. The two possibilities are $3[1,0,-2]^T-2[0,1,-3]^T$ and $-3[1,0,-2]^T+2[0,1,-3]^T$. The latter is equal to $[-3,2,0]^T$, so you’re done.
