# Basis of of subspaces of $R^3$

I'm having a little trouble with this question:

How do I find the basis of the set of vectors lying in the plane 2x − y − z = 0?

I'm stuck on how to start on this question

I tried to start by setting y= 2x-z

I'm not sure where to go from here

If we set $y=0$, then $x$ and $z$ satisfy the equation $2x-z=0$. Note that for every real number $\alpha$, the vector $$\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\alpha\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$$ satisfies this equation. Now set $z=0$, then for every real number $\beta$ the vector $$\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\beta\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$$ satisfies $2x-y=0$. If we add these vectors together, they satisfy the equation $2x-y-z=0$. Indeed: $$2x-y-z = 2(\alpha+\beta)-\alpha-\beta=0$$ So a basis of this plane is given by $$\mathcal{B} =\left\{\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix},\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}\right\}$$
Your start $y= 2x-z$ is good. It means that every vector $$\begin{pmatrix}x\\2x-z\\z\end{pmatrix}, \quad\text{x,y being real numbers,}$$ belongs to the plane. Hence every point in that plane may be expressed as $$\begin{pmatrix}x\\2x-z\\z\end{pmatrix}=x\begin{pmatrix}1\\2\\0\end{pmatrix}+ z\begin{pmatrix}0\\-2\\1\end{pmatrix}.$$