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