<s>[INST] Let $W = \textit{span}\{\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3\}$... Let $W = \text{span}\{\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3\}$, where \begin{equation*}
\mathbf{w}_1 = \begin{pmatrix}
1 \\
2 \\
1
\end{pmatrix}, \quad \mathbf{w}_2 = \begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix}, \quad \mathbf{w}_3 = \begin{pmatrix}
2 \\
2 \\
2
\end{pmatrix}.
\end{equation*}
1. Find a basis $\{\mathbf{v}_1,\ldots,\mathbf{v}_n\}$ for $W$.
Let
\begin{equation*}
A = \begin{pmatrix}
1 & 1 & 2 \\
2 & 0 & 2 \\
1 & 1 & 2
\end{pmatrix}
\end{equation*}
be a matrix that contains the vectors in $W$. Row reducing $A$ leads to $\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{pmatrix}$. We see that the first two columns are pivots (meaning they are linearly independent) so the basis for $W$ is
\begin{equation*}
\left\{\begin{pmatrix}
1 \\
2 \\
1
\end{pmatrix},\begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix}\right\}.
\end{equation*}
2. Find a basis $\{\mathbf{u}_1,\ldots,\mathbf{u}_m\}$ for $W^{\perp}$.
So do we transpose $A$ then reduce and then find a basis for the null space?
3. Write each $x\in \mathbb{R}^3$ uniquely in the form
\begin{equation*}
\mathbf{x} = a_1\mathbf{v}_1+\ldots+a_n\mathbf{v}_n+b_1\mathbf{u}_1+\ldots +b_m\mathbf{u}_m.
\end{equation*}
Here the coeffieients $a_j$ and $b_k$ will depend on $\mathbf{x}$.
Not sure how to do that one. Please help!
 [/INST] For $2)$, you could take the cross product of $(1,2,1)$ and $(1,0,1)$.
Get $\begin {vmatrix} i&j&k\\1&2&1\\1&0&1\end{vmatrix}=(2,0,-2)$.
For $3)$, given $\vec x=(x_1,x_2,x_3)$, we need to solve $\begin {pmatrix}1&1&2\\2&0&0\\1&1&-2\end{pmatrix}\vec y=\vec x$.  
Thus, $\vec y=\begin {pmatrix}1&1&2\\2&0&0\\1&1&-2\end{pmatrix}^{-1}\vec x$, where $\vec y=(a_1,a_2,b_1)$.
</s>