Answer
See answers below
Work Step by Step
We are given: $S=\{(x,y,z)\in R^3: x-3y+z=0\}$
Since $x-3y+z=0 \rightarrow z=-x+3y$ with $v=(x,y,z) \in S$ we have $ v=(x,y,-x+3y)=(x,0,-x)+(0,y,3y)=x(1,0,-1)+y(1,0,3)$.
Then $v \in$ span $\{ (1,0,-1);(0,1,3)\}$ and $S \subset $ span $\{ (1,0,-1);(0,1,3)\}$.
We can set: $v=a(1,0,-1)+b(1,0,3) \in$ span $\{ (1,0,-1);(1,0,3) \}$ then $v=(a,b,-a+3b)$ and then $v \in S$. Thus span $\{ (1,0,-1);(0,1,3)\} \subset S \rightarrow S=$span $\{ (1,0,-1);(0,1,3)\}$
The set of vectors is a linearly independent spanning set for $S$.
Hence $\{ (1,0,-1);(0,1,3)\}$ is a basis for $S$ and $dim S=2$
