Proving that 2 vectors exist that satisfy an equation I'm having trouble with the following problem: 

It gives a hint to specify the vectors with formulas involving a, b, this is where I have gotten to:
v1(a+b) = z
v2(a+b) = z

v1(a+b) - v2(a+b) = 0

(v1 - v2)(a+b) = 0

Now the above equation is a heterogeneous vector space, therefore [0,0,0] must be one of the vectors. Then I get stuck, what do I need to do next, am I even on the right track?
 A: To determine the value of $\,z\,$ in $\,z=ax+by$, we must choose one value for $\,x\,$ and one value for $\,y$.
Hence, $\,x,\,y\,$ are "free independent variables" for us to choose. In other words, we can generate the whole set of points satisfying the equation by choosing all the possible values of $\,x\,$ and $\,y$.
So how to choose all possible values for $\,x,\,y\,$?
Since $\,x,\,y\,$ are independent, we let $\,x=1\,$ while $\,y=0\,$ so that $\,y\,$ will no longer "affect" $\,x\,$. As a result, the first vector $\,v_1\,$ should be 
$$v_1=(1,0,a)$$
As you can see, if we multiply $\,v_1$ by any real constant, we can run through all the possibilities of $\,x$.
Next, we set $\,x=0\,$ and $\,y=1$, which gives the second vector $\,v_2\,$
$$v_2=(0,1,b)$$
This time, we can get all the possibilities of $\,y\,$ by multiplying $\,v_2\,$ by any real constant.
Together, the linear combinations of $\,v_1$ and $\,v_2$ will give us all the possibilities of $\,x\,$ and $\,y\,$. So the whole set of points satisfying the equation will be the set 
$$\lambda v_1+\mu v_2$$
where $\,\lambda,\,\mu\,$ are real constants.
( This result is reasonable because$\ \ \lambda v_1+\mu v_2=(\lambda,\,\mu,\,a\lambda+b\mu)\ $)
