= Solution
Subtracting twice the first equation from the second gives $(\lambda+2)z=0$. The third equation gives $x=1+\lambda y$, and substitution in the first then gives $z=(\lambda+1)y$. Consequently $(\lambda+2)(\lambda+1)y=0$. The complete real <solution set> of this <linear system> is
$$
\boxed{\begin{array}{c|c}
\lambda\notin\{-1,-2\}&(x,y,z)=(1,0,0)\\
\lambda=-1&(x,y,z)=(1-s,s,0),\quad s\in\mathbb R\\
\lambda=-2&(x,y,z)=(1-2s,s,-s),\quad s\in\mathbb R.
\end{array}}
$$
There is \b[no parameter value allowing both $x<0$ and $y<0$]. In either exceptional family, $y=s<0$ forces $x>0$; in the remaining case, $y=0$. The exceptional values must be handled before dividing by their factors.
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