Determine which vectors in $R^5$ belong to the span 
Determine which vectors in $R^5$ belong to both $span$ {$(a_{1}, b_{1}, c_{1}, d_{1}, e_{1}), (a_{2}, b_{2}, c_{2}, d_{2}, e_{2})$} and $span$ {$(f_{1}, g_{1}, h_{1}, i_{1}, j_{1}), (f_{2}, h_{2}, i_{2}, j_{2}, k_{2}), (f_{3}, g_{3}, h_{3},
 i_{3}, j_{3})$}.
where $a,b,c,d,e$ and $f,h,i,j,k$ are random vector components that I put as an example. 

Question: How do i actually find the vector in $R^5$ in both the spans? It would be great if someone could explain the concept behind this because I do not actually get what the question is asking for. 
My Attempt: 
My only guess of doing this is the following (thus i need guidance): 
$$
        \begin{pmatrix}
        a_{1} & b_{1} & c_{1} & d_{1} & e_{1} \\
        a_{2} & b_{2} & c_{2} & d_{2} & e_{2}\\
        \end{pmatrix}
$$
$$
        \begin{pmatrix}
        f_{1} & g_{1} & h_{1} & i_{1} & j_{1}\\
        f_{2} & g_{2} & h_{2} & i_{2} & j_{2}\\
        f_{3} & g_{3} & h_{3} & i_{3} & j_{3}\\
        \end{pmatrix}
$$
and reduce them in their row-echelon forms, even then I do not know how to proceed because I really do not understand the question. 
Thanks. 
 A: I think you can see how things like this work in a practile example. For simplicity, I will consider $\mathbb{R}^3$ being the vector space and I'll give the values of the vectors in the span, and the other cases will be quite similar.
"Determine which vectors in $\mathbb{R}^3$ belong to both $span\{(0,1,1),(1,-1,1)\}$ and $span\{(0,1,2),(1,0,1)\}$."
This is, find $v \in \mathbb{R}^3$ such that $v \in span\{(0,1,1),(1,-1,1)\} \cap span\{(0,1,2),(1,0,1)\}$ $\Rightarrow$ Exists elements $a_1,a_2,a_3,a_4$ on the field such that $v = a_1(0,1,1)+a_2(1,-1,1) = a_3(0,1,2)+a_4(1,0,1)$ (each equality is because v is in each span)  $\Rightarrow$ $(a_2,a_1-a_2,a_1+a_2) = (a_4,a_3,2a_3+a_4) \Rightarrow a_2=a_4, ~~ a_1 - a_2 = a_3,~~ a_1+a_2=2a_3+a_4$.
Now solve this for $a_1$ and $a_2$ (or $a_3$ and $a_4$) and we have $a_1=2a_3; ~~a_2=a_3$, replacing at $v$ equation get $$v = a_1(0,1,1)+a_2(1,-1,1) = 2a_3(0,1,1) + a_3(1,-1,1) = a_3(1,1,3) $$
Therefore, all $v$ that is in both spans is a linear combination of ${(1,1,3)}$. In other words, $span{(0,1,1),(1,−1,1)}∩span{(0,1,2),(1,0,1)}= span\{(1,1,3)\}$
