12. If $(x, y, z)$ is
$$
\left\{\begin{array}{l}
5 x-3 y+2 z=3, \\
2 x+4 y-z=7, \\
x-11 y+4 z=3
\end{array}\right.
$$

the value of $z$ is ( ).
(A) 0
(B) -1
(C) 1
(D) $\frac{4}{3}$
(E) $*$
12. (E).
(1) $-(3)$ yields
$4 x+8 y-2 z=0$. That is, $2 x+4 y-z=0$, which contradicts (2). Therefore, the system of equations has no solution.