## Elementary Linear Algebra 7th Edition

$S$ does not span $R^3$.
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 $S$.