Answer
the set $S=\{(6,-5),(12,-10)\}$ is not a basis for $R^2$.
Work Step by Step
The set $S=\{(6,-5),(12,-10)\}$ is not a basis for $R^2$ because the vectors in $S$ are not linearly independent. Because one can see that $(12,-10)=2(6,-5)$, and hence we have the non trivial combination
$$2(6,-5)-(12,-10)=(0,0)$$
Which means that the vectors $(6,-5),(12,-10)$ are not Linearly independent. Hence, the set $S=\{(6,-5),(12,-10)\}$ is not a basis for $R^2$