Let $\mathbf{x} = [x_1, x_2, x_3]^t$ the 'data' where $x_1$ is considered to be 'noise', $M$ a $3\times 3$-matrix with full rank, and $\mathbf{y} = M\mathbf{x}$ the obserced mixture.
Let $m^-_i$ denote the $i^\text{th}$ row of the inverse $M^-$ such that $m^-_i\mathbf{y} = x_i$.
Let $N$ be a $2 \times 3$-matrix whose rows span the orthogonal complement of the space spanned by $m^-_1$ (not necessarily consisting of $m^-_2, m^-_3$), i.e. the 'noise' $x_1$ is being neglected.
(When) Is it possible to find a vector $\mathbf{p} = [p_1, p_2]$ such that $\mathbf{p}N\mathbf{y} = \mathbf{p}NM\mathbf{x} = x_2$?
If, how does $\mathbf{p}$ relate to $m^-_2$?
Attempts
Ideally it must hold that $\mathbf{p}NM\mathbf{x} = m^-_2M\mathbf{x}$, hence $\mathbf{p}N=m^-_2$, which is a system of 3 equations and 2 unknowns, hence overdetermined and has no solution in general.
This somehow seems odd, as it intuitively should still be possible to get $x_2$ since only $x_1$ has been neglected..where am I fool messing up things?
Example
$M = \begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}$ such that $\mathbf{y} = \begin{pmatrix} x_1+x_2+x_3 \\ x_2+x_3 \\ x_3 \end{pmatrix}$ and $m^-_2 = [0, 1, -1]$.
Let $N = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ such that $N\mathbf{y} = \begin{pmatrix} x_1+2x_2+2x_3 \\ x_3 \end{pmatrix}$ from which we cannot obtain $x_1$.
The reason why things go wrong is that your matrix $M$ is not orthogogonal.
Let's look at your example a bit more closely: First of all the conclusion you draw in the end should probably be that we cannot obtain $x_1$ since here $x_2$ plays the role of your noise. What you do see in this example is that if you take the matrix $N$ you do not neglect $x_2$, as it still shows up in $Ny$. This is because taking the orthogonal complement of your vector $m_\bar 2 = [0, 1, -1]$ will not get rid of this component. However, if your measurement matrix $M$ was orthogonal this would work but then the vectors $m_\bar 1, m_\bar 3$ would span the orthogonal complement of $m_\bar 2$ anyways.
I hope this helps...