Answer
$S$ does not span $R^3$.
Work Step by Step
Since we have $(2,0,6)=2(1,0,3)$, then these vectors are linearly dependent. Now we can consider the combination
$$a(1,0,3)+b(2,0,-1)+c(4,0,5)=(0,0,0).$$
We have the system
\begin{align*}
a+2b+4c&=0\\
3a-b+5c&=0.
\end{align*}
The above system consists of two homogeneous equations in three variables then it has nontrivial solutions.
Hence, the vectors $(1,0,3),(2,0,-1),(4,0,5)$ are linearly dependent and since $R^3$ has the dimension $3$ then $S$ has less than three linearly independent vectors. Consequently, $S$ does not span $R^3$. Moreover, $S$ spans a plane generated by the vectors of $S$.
