 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 isThere is 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.