Find two vectors $a$ and $b$ in $M$ which are not scalar multiples of the other and show that $M$ is spanned by $a$ and $b$ 
Let $M$ be the subspace 
  $$M = \{ (x_1,x_2,x_3)\in \mathbb{R}^3 :\, x_1 - x_2 + 2x_3 = 0\}$$
  Find two vectors $a$ and $b$ in $M$ neither of which is a scalar multiple of the other. Then show that $M$ is the linear span of $a$ and $b$.

Here is my plan from end to start.
3) Show that $M$ is a linear span of two vectors, $a$ and $b$. (Does this mean that $M$ is a two-dimensional space because its spanned by two vectors?)
2) To do that, I must find two vectors $a$ and $b$ in $M$ that are not scalar multiples of each other.
1) Starting with the definition of $M$, one vector is already provided, $a = \langle ~x_1, -x_2, 2x_3~\rangle$.

I am struggling on finding a second vector, $b$, that is not a scalar multiple of $a$. I considered the zero vector and 1d or 2d vectors, but found counter examples so this was the wrong approach. 
Then I thought, how would I determine if $x_1 - x_2 + 2x_3$ spanned $M$ alone? Maybe by showing that $M$ can't be spanned by a single vector, then the second vector needed would be more apparent, but I don't know how I would make such a determination.
 A: HINT
Note that if $\vec{x} \in M$ then its components satisfy $$x_1-x_2+2x_3 = 0 \iff x_1 = x_2-2x_3.$$ Now we have
$$
\vec{x}
 = \begin{pmatrix} x_1 \\ x_2 \\x_3 \end{pmatrix}
 = \begin{pmatrix} x_2-2x_3 \\ x_2 \\x_3 \end{pmatrix}
 = x_2 \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}
 + x_3 \begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}
$$
Can you complete the problem now?
A: Are you not clear on the definition of "span"?  gt6989b showed how any vector in this space can be written as a linear combination of those two vectors.  Yes, that is exactly what "span" means!
You also ask "(Does this mean that M is a two-dimensional space because its spanned by two vectors?)".  NO, just spanning is not sufficient.  For example, The space of vectors (x, 0, 0) can be "spanned" by {(1, 0, 0), (2, 0, 0), (3, 0, 0)}.
In order to conclude that M is two dimensional, you would need to determine that there is a basis consisting of two vectors.  To be a "basis" the vectors must span the space and be independent.  In this problem, it is fairly easy to show that two vectors such that one is NOT a multiple of the other ARE independent.
