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.
 A: Hint: That is the equation of line that goes through point (1,1,0), with vector (1,2,1). ¿Can you find two planes that intersect in that line? (that means the number of solutions to that will be infinite)
A: We have $A\begin{pmatrix}
1\\ 
1\\
0\\
\end{pmatrix}$ Now choose any matrix with null space as 
$w\begin{pmatrix}
1\\ 
1\\
0\\
\end{pmatrix}$ and we get the answer(Just multiply to find $b$).
A: 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.
