[Edit: Removed most of the original version, as it was ill-formed and actually contained errors. A downvote is due!]
$Mw=0$, so $w$ is an eigenvector belonging to eigenvalue $\lambda=0$. As $M$ is a 3x3 matrix, it has at most two other eigenvalues (counted with multiplicity).
Any eigenvector belonging to a non-zero eigenvalue must lie in $V=<v>^\perp$, because its image lies in that plane.
A fixed point ($\neq0$) is an eigenvector belonging to eigenvalue $\lambda=1$, and by the previous point $\in V$.
The restriction $M|_V$ of $M$ onto the plan $V$ is a mapping $V\rightarrow V,$
$\lambda=1$ may be a double root of the characteristic equation of $M|_V$, but the corresponding eigenspace may have dimension one only. In that case we can say that geometrically $M|_V$ is a shearing (see http://en.wikipedia.org/wiki/Transformation_matrix#Shearing).
There is no reason to think that $M|_V$ would
have any real eigenvalues, but the fixed points come from there, so for that reason $M|_V$ and its potential for having eigenvalue $\lambda=1$ received the special attention in the previous paragraph.
It is, of course, possible that $w\in V$. There is no reason to think that $w$ would be ortohogonal to $V$, unless we can choose $w=v$. (therefore we cannot conclude that $M$ would be an orthogonal projection at this point as I errorneously stated in the erased version).