Answer
$S$ is a basis for $R^3$.
Work Step by Step
The set $S=\{(1,5,3),(0,1,2),(0,0,6)\}$ is a linearly independent set of vectors. Indeed,
assume that
$$a(1,5,3)+b(0,1,2)+c(0,0,6)=(0,0,0), \quad a,b,c\in R.$$
Then, we have the following system of equations
\begin{align*}
a&=0\\
5a+b&=0\\
3a+2b+6c&=0.
\end{align*}
Solving the above equations, we find that $a=0,b=0,c=0$ and hence $S$ is linearly independent. Now, since $R^3$ is vector space of dimension $2$, then by Theorem 4.12, $S$ is a basis for $R^3$.
