Let's determine all triples of numbers \(a, b, c\) for which the following equation is an identity:

$$
|a x+b y+c z|+|b x+c y+a z|+|c x+a y+b z|=|x|+|y|+|z| \text {. }
$$
If the above equation is an identity, then substituting any value for $x, y$, and $z$ will yield an equality. If $x=1$, $y=z=0$, then

$$
|a|+|b|+|c|=1
$$

if, however, $x=y=z=1$, then

$$
|a+b+c|=1
$$

follows. In the inequality $|a+b+c| \leqq|a|+|b|+|c|$, which is always true, equality holds. It is known that this is precisely the case when $a, b$, and $c$ have the same sign.

If $x=1, y=-1, z=0$, then the following condition is obtained:

$$
|a-b|+|b-c|+|a-c|=2
$$

Notice that in (1), (2), and (3), the roles of $a, b$, and $c$ are interchangeable, so when examining these three conditions, it can be assumed that $a \geqq b \geqq c$. Taking this into account, in (3) the absolute value signs can already be omitted, and $a-c=1$ follows.

If neither $a$ nor $c$ is 0, then, being of the same sign, their difference can only be 1 if one of them has an absolute value greater than 1. However, this is impossible due to (1). Therefore, either $a=0$ or $c=0$. If $a=0$, then $c=-1$ and thus, by (1), $b=0$. If, however, $c=0$, then $a=1$, and $b$ is also 0 in this case. We have thus found that the solution is essentially unique: among the numbers $a, b, c$, two are equal to 0, and the third has a value of either +1 or -1.