194. For what real $a, b, c$ is the equality $|a x+b y+c z|+|b x+c y+a z|+|c x+a y+b z|=$

$$
=|x|+|y|+|z|
$$

true for all real $x, y, z$?
194. Answer: two of the numbers $a, b, c$ are 0, the third is $\pm 1$ (a total of 6 solutions).

Substituting into the condition sequentially: $x=y=z=1$, then $x=y=0, z=1$, and finally $x=1, y=-1, z=0$, we get the system $|a+b+c|=1,|a|+|b|+|c|=1,|a-b|+$ $+|b-c|+|c-a|=2$. Since $|a+b+c|=|a|+|b|+$ $+|c|$, all numbers $a, b, c$ have the same sign, i.e., $a b \geqslant 0, b c \geqslant 0$, $a c \geqslant 0$. However, in the inequality

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

(derived from $|a-b| \leqslant|a|+|b|$ and two similar inequalities), equality is possible only when $a b \leqslant 0, b c \leqslant 0, a c \leqslant 0$. Therefore, $a b=b c=a c=0$, so two of the numbers $a, b, c$ are 0, and the third (as seen from any of the three equations) is 1 in absolute value.