Answer
No basis for $R^3$
Work Step by Step
We are given:
$\{(1,-1,1),(2,5,-2),(3,11,-5)\}$
We can see that $(-1)(1,-1,1)+2(2,5,-2)=(-1,1,-1)+(4,10,-4)=(3,11,-5)$
Hence the set $\{(1,-1,1),(2,5,-2),(3,11,-5)\}$ is linearly independent in $R^3$.
Therefore, there is no basis for $R^3$
