Base of Subspace with vectors Let E be the vector subspace of $R^3$ generated by it vectors $v1 = (1,2,0)$ and $v2 = (-1,0,2)$
How can find a basis of E between the following vectors?
$$w1=(-2,-12,8), w2=(-12,-2,-8), w3=(-2,-3,1), w4=(1,-1,-3), w5=(6,3,-9), w6=(0,1,1)$$
I know that I have to demonstrate some effort solving the exercise first. I'm really lost with vectors and subspaces.
I tried $$a(1,2,0) + b(-1,0,2) = w1=(-2,-12,8)$$
$a = -6$
$b = 4$
$a-b=-2$ But this is not satisfied, thus $w1$ is not a basis.
I wonder if I'm in the right path, which are the conditions with the right basis? 
I have found that with $w3,w4,w6$ the three above equations are satisfied. Now what can I do with this?

None of the $w_i$ is a multiple of another. So if you can find two of them in the space generated by $v_1$ and $v_2$, you will be finished. This can be attempted like you started. The fact that one of the last two coordinates of the $v_i$ is $0$ makes the job easier.
We deal first with $w_1$. If $w_1=av_1+bv_2$, then from the last two coordinates we see that we must have $a=-6$ and $b=4$. Then the first coordinate of $av_1+bv_2$ would be $-10$. But this is not the first coordinate of $w_1$, so $w_1$ is eliminated.
Let's look at $w_6$. Looking at the last two coordinates, if this is to be $av_1+bv_2$ we need $a=1/2$ and $b=1/2$. That works, for it produces the right first coordinate of $w_6$. 
Since $w_6=(1/2)v_1+ (1/2)v_2$, it is in the space spanned by $v_1$ and $v_2$.
Thus $w_1$ is no good, and $w_6$ is good. Four left to examine!
