33. (MON 4) In what case does the system \[  $$ \begin{aligned} & x+y+m z=a, \\ & x+m y+z=b, \\ & m x+y+z=c, \end{aligned} $$ \]  have a solution? Find the conditions under which the unique solution of the above system is an arithmetic progression.
33. If $m \notin\{-2,1\}$, the system has the unique solution  
$$ x=\frac{b+a-(1+m) c}{(2+m)(1-m)}, \quad y=\frac{a+c-(1+m) b}{(2+m)(1-m)}, \quad z=\frac{b+c-(1+m) a}{(2+m)(1-m)} $$  
The numbers $x, y, z$ form an arithmetic progression if and only if $a, b, c$ do so. For $m=1$ the system has a solution if and only if $a=b=c$, while for $m=-2$ it has a solution if and only if $a+b+c=0$. In both these cases it has infinitely many solutions.