Answer
See answers below
Work Step by Step
We are given: $S=\{v \in R^3: v=(r,r-2s,3s-5r)\}$ with $r,s \in R$
Assume that $v=(r,r-2s,3s-5r) \in S$ we have $(r,r-2s,3s-5r)=(r,r,-5r)+(0,-2s,3s)=r(1,1,-5)+s(0,-2,3)$.
Then $v \in$ span $\{ (1,1,-5);(0,-2,3)\}$
We can set: $v=r(1,1,-5)+s(0,-2,3) \in$ span $\{ (1,1,-5);(0,-2,3)\}$ then $v=(r,r-2s,3s-5r)$ and then $v \in S$. Thus span $\{ (1,1,-5);(0,-2,3)\} \subset S \rightarrow S=$span $\{ (1,1,-5);(0,-2,3)\}$
$\{ (1,1,-5);(0,-2,3)\}$ is a linearly independent spanning set for $S$.
Hence $\{ (1,1,-5);(0,-2,3)\}$ is a basis for $S$ and $dim S=2$
