Solve the following system of equations ( $t$ is a real parameter):

$$
t \cdot(x+y+z)=0, t \cdot(x+y)+z=1, t \cdot x+y+z=2 .
$$
I. If $t=0$, then from the second equation $z=1$. Substituting this into equation (3) gives $y=1$, and the value of $x$ can be any number.

II. If $t \neq 0$, from the first equation $t(x+y)+t z=0$, and from the second equation $t(x+y)=1-z$, hence $z=-\frac{1}{t-1}$, assuming that $t \neq 1$. (This case needs to be examined separately.)

From the difference of equations (2) and (3), we get $y=-\frac{1}{t-1}$.

Finally, from equation (3), $x=\frac{2}{t-1}$.

By substitution, it can be verified that $x=\frac{2}{t-1}, y=z=-\frac{1}{t-1}$ is indeed a solution to the system of equations.

III. Finally, if $t=1$, from equation (1) $x+y+z=0$, and from the second equation $x+y+z=1$.

This is a contradiction, so the system of equations has no solution in this case.

Note. Very few examined the last case. These solutions did not receive full marks due to the incomplete solution.