1. For any integers $x, y, z$, all real numbers $a, b, c$ that satisfy the equation $|a x+b y+c z|+|b x+c y+a z|+|c x+a y+b z|=| x|+| y|+| z|$ are $\qquad$.
$$
\begin{array}{r}
\text { = }, 1 .(a, b, c)=(1,0,0),(-1,0,0), \\
(0,1,0),(0,-1,0),(0,0,1),(0,0,-1)+
\end{array}
$$

Group.
$$
\begin{array}{l}
\text { Let } x=1, y=z=0, \text { then } \\
|a|+|b|+|c|=1 .
\end{array}
$$
$$
\text { Let } x=y=z=1 \text {, then }|a+b+c|=1 \text {. }
$$

Thus, $|a+b+c|=|a|+|b|+|c|, a, b, c$ have the same sign.
Let $x=y=1, z=-1$, then
$$
|a+b-c|+|b+c-a|+|c+a-b|=3 \text {. }
$$

By $a, b, c$ having the same sign, then
$$
\begin{array}{l}
|a+b-c| \leqslant|a+b+c|, \\
|b+c-a| \leqslant|a+b+c|, \\
|c+a-b| \leqslant|a+b+c| .
\end{array}
$$

Therefore, $|a+b-c|=|b+c-a|$
$$
=|c+a-b|=|a+b+c|=1 .
$$
$a, b, c$ must have two zeros.