$4 \cdot 61$ For what real values of $a$, $b$, and $c$, does the equation
$$
\begin{aligned}
& |a x+b y+c z|+|b x+c y+a z|+|c x+a y+b z| \\
= & |x|+|y|+|z|
\end{aligned}
$$

hold for all real numbers $x, y, z$?
[Solution] Let $x=y=z=1$, we get
$$
|a+b+c|=1 \text {, }
$$

Let $x=y=0, z=1$, we get
$$
|a|+|b|+|c|=1 \text {, }
$$

Let $x=1, y=-1, z=0$, we get
$$
|a-b|+|b-c|+|c-a|=2 .
$$

From (1) and (2), we know that, except for 0, the signs of $a, b, c$ are the same, i.e.,
$$
a b \geqslant 0, b c \geqslant 0, c a \geqslant 0 .
$$

Since
$$
\begin{aligned}
& |a-b|+|b-c|+|c-a| \\
\leqslant & (|a|+|b|)+(|b|+|c|)+(|c|+|a|),
\end{aligned}
$$

Therefore, from (2) we get
$$
|a-b|+|b-c|+|c-a| \leqslant 2 .
$$

From (3), we know that the equality holds in the above inequality, so we must have
$$
\left\{\begin{array}{l}
|a-b|=|a|+|b|, \\
|b-c|=|b|+|c|, \\
|c-a|=|c|+|a| .
\end{array}\right.
$$

From the above, we know
$$
a b \leqslant 0, b c \leqslant 0, c a \leqslant 0,
$$

From (4) and (7), we get
$$
a b=0, b c=0, c a=0,
$$

From (8), we know that at least two of $a, b, c$ are 0, and from (1), we get that two of $a, b, c$ must be 0, and the third one is $\pm 1$.
Therefore, in the following six cases, the given equations hold for all real numbers $x, y, z$:
$$
\left\{\begin{array} { l } 
{ x = 0 , } \\
{ y = 0 , } \\
{ z = 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x = 0 , } \\
{ y = 0 , } \\
{ z = - 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x = 1 , } \\
{ y = 0 , } \\
{ z = 0 ; }
\end{array} \left\{\begin{array} { l } 
{ x = - 1 , } \\
{ y = 0 , } \\
{ z = 0 ; }
\end{array} \left\{\begin{array} { l } 
{ x = 0 , } \\
{ y = 1 , } \\
{ z = 0 ; }
\end{array} \left\{\begin{array}{l}
x=0 \\
y=-1 \\
z=0
\end{array}\right.\right.\right.\right.\right.\right.
$$