Prove S is a subspace of $R^3$ Let S={(a,b,c)$\in$$R^3$|2a+b-3c=0}. Prove S is a subspace of $R^3$
I just asked another question like this earlier and this is just to confirm im on the right track when approaching problems like these. After starting this problem, i am finding myself confused when it comes to making sure its closed under addition(so far). 
so far i think i have proved the set is not empty with: 
let 0=(0,0,0),
2(0)+0-3(c)=0 therefore, set is not empty. 
let u=($a_1,b_1,c_1$) and w=($a_2,b_2,c_2$). 
u+w= $(a_1+a_2),(b_1+b_2),(c_1+c_2) $
At this point, where does the definition "2a+b-3c=0" come into play? How do i proceed from here to show closure under addition?

If we suppose $u=(u_1,u_2,u_3)$ and $w=(w_1w_2,w_3)$ are arbitrary vectors in $S$, so what we are saying is that
$2u_1+u_2-3u_3=0$ and that $2w_1+w_2-3w_3=0$. Now, we see that the vector
$$u+w=(u_1+w_1,u_2+w_2,u_3+w_3)$$
satisfy 
$$2(u_1+w_1)+(u_2+w_2)-3(u_3+w_3)$$
$$\begin{align}  
&= (2u_1+u_2-3u_3)+(2w_1+w_2-3w_3) \\
&= 0+0 \\
&= 0
\end{align}$$
that is, the vector $u+w$ also belongs in $S$. Do the same thing with the vector $cu$, where $c$ is an arbitrary scalar.
