For which numbers $a, b, c$ will the following identity hold:

$$
|a x+b y+c z|+|b x+c y+a z|+|c x+a y+b z|=|x|+|y|+|z|
$$
Let's first consider the expression under investigation for the values $x=y=z=1$:

$$
3|a+b+c|=3
$$

which means

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

Then, by substituting $x=y=0, z=1$,

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

Comparing (1) and (2), and squaring both sides,

$$
a^{2}+b^{2}+c^{2}+2|a b|+2|a c|+2|b c|=a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 b c
$$

which simplifies to

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

This precisely means that

$$
a b \geqslant 0, \quad a c \geqslant 0, \quad b c \geqslant 0
$$

Now let $x=1, y=-1, z=0$. Then,

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

Using the fact that for any numbers $A, B$, $|A|+|B| \geqslant|A-B|$, and also considering (2),

$$
2=|a-b|+|b-c|+|c-a| \leq|a|+|b|+|b|+|c|+|c|+|a|=2(|a|+|b|+|c|)=2
$$

This can only hold if

$$
|a-b|=|a|+|b|, \quad|b-c|=|b|+|c|, \quad|c-a|=|c|+|a|
$$

Squaring the first equation,

$$
a^{2}+b^{2}-2 a b=a^{2}+b^{2}+2|a b|
$$

which implies

$$
|a b|=-a b
$$

The same can be done for the other two cases, leading to

$$
a b \leqslant 0, \quad b c \leqslant 0, \quad c a \leqslant 0
$$

This, together with (3), means that

$$
a b=b c=c a=0
$$

which holds if and only if at least two of the numbers $a, b, c$ are 0.

Considering the case $a=b=0$,

$$
|c| \cdot|z|+|c| \cdot|x|+|c| \cdot|y|=|x|+|y|+|z|
$$

we get $c= \pm 1$. Similarly, for $a=c=0$, $b= \pm 1$, and for $b=c=0$, $a= \pm 1$. These are indeed valid solutions, so the solutions are given by the following triples:

$(0,0,1),(0,0,-1),(0,1,0),(0,-1,0),(1,0,0),(-1,0,0)$.

Based on the work of Márton Hegedüs (Főv. Fazekas M. Gyak. Gimn., I. o. t.)