Finding common vectors in $R^5$ of both spans 
Determine which vectors in $R^5$ belong to both $sp${$(1, 0, 3, 5, 1),
 (0, 2, 0, 1, 5)$} and $sp${$(1, 0, 2, 5, 1), (1, 3, 1, 1, 0), (2, 7,
 0, 0, 1)$}.

My approach:
I've interpreted this question as a particular vector $v$ $ϵ$ $R^5$, such that 
$v$ $ϵ$ $sp${$(1, 0, 3, 5, 1),
 (0, 2, 0, 1, 5)$} $∩$ $sp${$(1, 0, 2, 5, 1), (1, 3, 1, 1, 0), (2, 7,
 0, 0, 1)$} 
which exists real number $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. 
$\implies$ $v = a_{1} (1, 0, 3, 5, 1) + a_{2} (0, 2, 0, 1, 5) = a_{3} (1, 0, 2, 5, 1) + a_{4} (1, 3, 1, 1, 0) + a_{5} (2, 7, 0, 0, 1)$
$\implies$ $$(a_{1}, 2a_{2}, 3a_{1}, 5a_{1} + a_{2}, a_{1} + 5a_{2}) = (a_{3} + a_{4} + 2a_{5}, 3a_{4} + 7a_{5}, 2a_{3} + a_{4}, 5a_{3} + a_{4}, a_{3} + a_{5})$$
$\implies$ 
$a_{1} = a_{3} + a_{4} + 2a_{5}$
$2a_{2} = 3a_{4} + 7a_{5}$
$3a_{1} =  2a_{3} + a_{4}$
$5a_{1} + a_{2} = 5a_{3} + a_{4}$
$a_{1} + 5a_{2} = a_{3} + a_{5}$
I understand the need to find a single relationship, so that I can list the linear combinations of the vectors that can span both. I'm stuck here, or am I doing it wrong? 

Hint: You're having a system of equations with $5$ variables. Bring it back to matrix $\mathbf{Ax}=0$ and solve it. The answer will be all vectors $\alpha \cdot \mathbf{x}$ where $\alpha \in \mathbf{R}$.
