Find a $2$ by $3$ system $Ax=b$ whose general solution is.... I came across a problem in my Linear Algebra book that says:  

Find  a  $2$ by $3$ system $Ax=b$ whose general solution is $x=\begin{pmatrix}
1\\ 
1\\ 
0
\end{pmatrix}+w\begin{pmatrix}
1\\ 
2\\ 
1
\end{pmatrix}.$  

The answer is given to be: $\begin{pmatrix}
1 &0  &-1 \\ 
 0&1  &-2 
\end{pmatrix}\begin{pmatrix}
u\\ 
v\\ 
w
\end{pmatrix}=\begin{pmatrix}
1\\ 
1\\ 
0
\end{pmatrix}$.   
I have no idea how it came. Can someone point me in the right direction? Thanks in advance for your time.
EDIT: There must be a typo on the part of the answer as the right hand part of the answer should be $2 \times 1$ matrix. But, how can I tackle it if I left aside the erroneous part of the answer.

Take a $2\times 3$ matrix: $\begin{pmatrix}
a &b  &c \\ 
 d&e  &f 
\end{pmatrix}$
And determine $a,\dots ,f$ such that $\begin{pmatrix}
v_1\\ 
v_2\\
v_3
\end{pmatrix}$(2nd $2\times1$ vector on the right) belongs to its null space.
Let $x=\begin{pmatrix}
u_1\\ 
u_2\\
u_3
\end{pmatrix}+w\begin{pmatrix}
v_1\\ 
v_2\\
v_3
\end{pmatrix}$
Then $b=Ax=A\begin{pmatrix}
u_1\\ 
u_2\\
u_3
\end{pmatrix}+wA\begin{pmatrix}
v_1\\ 
v_2\\
v_3
\end{pmatrix}$
As this is a general solution it means that we will be able to get all the solutions by varying $w\in R$.
As $b\in R^2$ is fixed then w cant vary unless $A\begin{pmatrix}
v_1\\ 
v_2\\
v_3
\end{pmatrix}=0$
This is the complete arguement.
