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?

$$2a_1+b_1-3c_1=0$$ $$2a_2+b_2-3c_2=0$$
Thus $2(a_1+a_2)+(b_1+b_2)-3(c_1+c_2)=0+0=0$
So $u+v \in S$
Similarly prove that $au \in S ,\forall a \in \Bbb{R}$
