Finding all the solutions of the system I'm struggling with the following problem:
Consider the system of equations of three unknowns $Ax=b$, for which we know that:
a) $A≠0$.
b) $(1, 2, 2)$ and $(0, 1, 1)$ are solutions,
c) $(0, a, b)$ is a solution if and only if $a=b=1$.
I'm asked to find all the solutions of the system.

The issue is that I'm not sure how to use c). Particularly the fact that no other solution of the form $(0, a, b)$ exists.
Also, the teacher uses $b$ first as a vector, and then as a scalar. My guess is that when $b$ is used as a vector it is meant to be $(b, b, b)$.
I'd appreciate if anyone could point me in the right direction.

I'm including the original problem here by petition of a user, it is however in Catalan.

I admit the teacher isn't clear in what is wanted, I believe he means us to solve
$$A\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} b \\ b \\ b \end{pmatrix}$$
such that for some fixed value $b=b_1$ we have that $(1, 2, 2)$ is a solution (similarly, for some fixed value $b=b_2$, we have that $(0,1,1)$ is a solution, in which case we know by c) that $b_2=1$), and such that $ay+bz=b$ if and only if $a=b=1$.
 A: (I only presume that a system whose coefficient matrix is $A$ has solutions satisfying the conditions (a), (b), and (c) regarding that "$AX=b$"-thing for a typo otherwise the problem would be senseless.)
Clearly the solution set's dimension is neither zero nor three.
Assuming the solution set has dimension $1$ (and assuming that $A$ has three columns, of course) the solution is a straight line given by
$$x=w+s\cdot u$$
with a suitable replacement-vector $w$ and a direction-vector $u$, where $u$ is an element of the kernel of $A$.  Now a candidate for $u$ is easily found: as $(1,2,2)$ and $(0,1,1)$ are solutions their difference $u=(1,1,1)$ is an element of the kernel.
For the replacement vector simply choose $w=(0,1,1)$; the solutions are given by
$$
\begin{pmatrix}
0\\1\\1
\end{pmatrix}
+s\begin{pmatrix}
1\\1\\1
\end{pmatrix},
$$
which may also be written as
$$
\begin{pmatrix}
-1\\0\\0
\end{pmatrix}
+r\begin{pmatrix}
-1\\-1\\-1
\end{pmatrix}.
$$
As $u$ and $w$ are not parallel the only solution who's first coordinate is zero is just $w$ as
$$\begin{pmatrix}0\\a\\b
\end{pmatrix}=\begin{pmatrix}0\\1\\1
\end{pmatrix}+s\begin{pmatrix}1\\1\\1
\end{pmatrix}$$
has only solution $s=0$.
In case the solution set's dimensions equals two, rref has the form
$$\begin{array}{ccc|c}
1 & \alpha & \beta & \gamma
\end{array}$$
hence the solutions are given by
$$
\begin{pmatrix}
\gamma\\0\\0
\end{pmatrix}
+r\begin{pmatrix}
\alpha\\-1\\0
\end{pmatrix}
+s\begin{pmatrix}
\beta\\0\\-1
\end{pmatrix}.
$$
As $(0,1,1)$ must be a solution we conclude that in this case $r=s=-1$ and hence $\gamma=\alpha+\beta$.
As $(1,2,2)$ must be a solution we conclude that in this case $r=s=-2$ and from here $\alpha+\beta=-1$, that is $\beta=-1-\alpha$.
Finally the solutions are given by
$$\begin{pmatrix}-1\\0\\0
\end{pmatrix}+s\begin{pmatrix}\alpha\\-1\\0
\end{pmatrix}+r\begin{pmatrix}-1-\alpha\\0\\-1\end{pmatrix}.$$
As obviously
$$\begin{pmatrix}-1\\0\\0
\end{pmatrix},\begin{pmatrix}\alpha\\-1\\0
\end{pmatrix}\text{ and} \begin{pmatrix}-1-\alpha\\0\\-1\end{pmatrix}$$
are linearly independent, $(0,a,b)$ is a solution iff $a=b=1$.
