Answer
See below
Work Step by Step
Given the set of vectors $\{(-4,1,3),(5,1,6),(6,0,2)\}$
Obtain: $\begin{vmatrix}
-4 & 5 & 6\\ 5 & 1 & 6 \\ 6 & 0 &2
\end{vmatrix}=-4.\begin{vmatrix}
1 & 0 \\ 6 &2
\end{vmatrix}-5.\begin{vmatrix}
1 & 0 \\ 3 &2
\end{vmatrix}+6.\begin{vmatrix}
1 & 1\\3 & 6
\end{vmatrix}=-4.2-5.2+6.3=0$
Thus, the set of vectors are coplanar and therefore set $\{(-4,1,3),(5,1,6),(6,0,2)\}$ does not span $R^3$
We can write set $S$ as $S=\{(x,y,z)\in M_R^3: x+13y-3z=0\}$
We have $x+13y-3z=0 \rightarrow x=3z-13y \forall (x,y,z) \in S$
The elements now can be written as:
$(x,y,z)=(3z-13y,y,z)=y(-13,1,0)+z(3,0,1)$
Notice that:
$(-13,1,0)=2(-4,1,3)-(5,1,6)\\
(3,0,1)=\frac{1}{2}(6,0,2)$
Substitute:
$y(-13,1,0)+z(3,0,1)\\
=y[2(-4,1,3)-(5,1,6)]+z.\frac{1}{2}(6,0,2)\\
=2y(-4,1,3)-y(5,1,6)+\frac{1}{2}(6,0,2)$
Every vector $(x,y,z) \in S$ can be written as a linear combination of vectors $\{(-4,1,3),(5,1,6),(6,0,2)\}$
Hence, the set of vectors $\{(-4,1,3),(5,1,6),(6,0,2)\}$ span subspace $S$