We want to first find some relations of linear dependence on $\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4,\,\vect{v}_5}$ that will allow us to “kick out” some vectors, in the spirit of Example SCAD. To find relations of linear dependence, we formulate a matrix $A$ whose columns are $\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4,\,\vect{v}_5$. Then we consider the homogeneous system of equations $\homosystem{A}$ by row-reducing its coefficient matrix (remember that if we formulated the augmented matrix we would just add a column of zeros). After row-reducing, we obtain
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 0 & 2 & -1\\
0 & \leading{1} & 0 & 1 & -2\\
0 & 0 & \leading{1} & 0 & 0
\end{bmatrix}
\end{equation*}
From this we find that solutions can be obtained employing the free variables $x_4$ and $x_5$. With appropriate choices we will be able to conclude that vectors $\vect{v}_4$ and $\vect{v}_5$ are unnecessary for creating
$W$ via a span. By Theorem SLSLC the choice of free variables below lead to solutions and linear combinations, which are then rearranged.
\begin{align*}
x_4=1, x_5=0&&\Rightarrow&&(-2)\vect{v}_1+(-1)\vect{v}_2+(0)\vect{v}_3+(1)\vect{v}_4+(0)\vect{v}_5=\zerovector&&\Rightarrow&&\vect{v}_4=2\vect{v}_1+\vect{v}_2\\
x_4=0, x_5=1&&\Rightarrow&&(1)\vect{v}_1+(2)\vect{v}_2+(0)\vect{v}_3+(0)\vect{v}_4+(1)\vect{v}_5=\zerovector&&\Rightarrow&&\vect{v}_5=-\vect{v}_1-2\vect{v}_2\\
\end{align*}
Since $\vect{v}_4$ and $\vect{v}_5$ can be expressed as linear combinations of $\vect{v}_1$ and $\vect{v}_2$ we can say that $\vect{v}_4$ and $\vect{v}_5$ are not needed for the linear combinations used to build $W$ (a claim that we could establish carefully with a pair of set equality arguments). Thus
\begin{equation*}
W=\spn{\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}}=\spn{\set{\colvector{2\\1\\1},\,\colvector{-1\\-1\\1},\,\colvector{1\\2\\3}}}
\end{equation*}
That the $\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}$ is linearly independent set can be established quickly with Theorem LIVRN.
There are other answers to this question, but notice that any nontrivial linear combination of $\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4,\,\vect{v}_5$ will have a zero coefficient on $\vect{v}_3$, so this vector can never be eliminated from the set used to build the span.