I think that you misunderstand what the quotient space is. Let $V=\operatorname{span}W_3$; as you said, $V$ is the set of all vectors in $\Bbb R^3$ of the form $\langle b-a,a,b\rangle$, where $a,b\in\Bbb R$. The elements of $\Bbb R^3/V$ are sets of vectors in $\Bbb R^3$: specifically, they are the sets $\langle x,y,z\rangle+V\;,$ where $\langle x,y,z\rangle$ ranges over all vectors in $\Bbb R^3$. For example, $\langle 0,1,2\rangle+V$ is one of the vectors in $\Bbb R^3/V$, and $\langle 0,3,4\rangle+V$ is another. But despite appearance, they are not distinct elements of $\Bbb R^3/V$: $\langle 0,1,2\rangle+V=\langle 0,3,4\rangle+V\;.$
How can I tell? $\langle 0,1,2\rangle+V$ is the set of all vectors in $\Bbb R^3$ of the form $\langle 0,1,2\rangle+\langle x,y,z\rangle$, where $\langle x,y,z\rangle\in V$. In other words, $\langle 0,1,2\rangle+V$ is the set of all vectors in $\Bbb R^3$ of the form $\langle 0,1,2\rangle+\langle b-a,a,b\rangle=\langle b-1,a+1,b+2\rangle\tag{1}$ for $a,b\in\Bbb R$. Is $\langle 0,3,4\rangle$ of the form $(1)$? Yes, with $a=b=2$. Thus, $\langle 0,3,4\rangle\in\langle 0,1,2\rangle+V\;,$ and it follows from basic properties of quotients that $\langle 0,1,2\rangle+V=\langle 0,3,4\rangle+V\;.\tag{2}$
A basic result that is useful here:
$\qquad\qquad\langle u,v,w\rangle+V=\langle x,y,z\rangle+V$ if and only if $\langle x,y,z\rangle-\langle u,v,w\rangle\in V$.
In my example, for instance, $\langle 0,3,4\rangle-\langle 0,1,2\rangle=\langle 0,2,2\rangle\in V$, so I knew right away that $(2)$ was true.
The problem requires you to find two distinct elements of $\Bbb R^3/V$. To do this, you must find $\langle u,v,w\rangle,\langle x,y,z\rangle\in\Bbb R^3$ such that $\langle u,v,w\rangle+V\ne\langle x,y,z\rangle+V$. This means ensuring that $\langle x,y,z\rangle-\langle u,v,w\rangle\notin V$. One easy way to start is to let $\langle u,v,w\rangle=\langle 0,0,0\rangle$, so that $\langle u,v,w\rangle+V=V$. Now just pick a suitable $\langle x,y,z\rangle$ and let your other element be $\langle x,y,z\rangle+V$.