10. Find the number of triples $(a, b, c)$ of real numbers such that the equation $|a x+b y+c z|+|b x+c y+a z|+|c x+a y+b z|=|x|+|y|+|z|$ holds for all real numbers $x, y, z$.
(1 mark)
Find the number of triples $(a, b, c)$ of real numbers such that the equation $|a x+b y+c z|+|b x+c y+a z|+|c x+a y+b z|=|x|+|y|+|z|$ holds for all real numbers $x, y, z$.
10. 6
10. Setting \(x=y=z=1\), we have \(|a+b+c|=1\).
Setting \(x=1\) and \(y=z=0\), we have \(|a|+|b|+|c|=1\).
Setting \((x, y, z)=(1,-1,0)\), we have \(|a-b|+|b-c|+|c-a|=2\).
Since \(|a+b+c|=|a|+|b|+|c|\), \(a, b, c\) must be of the same sign (unless some of them is/are equal to 0). But
\[
2=|a-b|+|b-c|+|c-a| \leq (|a|+|b|)+(|b|+|c|)+(|c|+|a|)=2(|a|+|b|+|c|)=2.
\]

The equality \(|a-b|=|a|+|b|\) holds only if \(a\) and \(b\) are of opposite signs (unless some of them is/are equal to 0). Similarly, \(b\) and \(c\) have opposite signs, and \(c\) and \(a\) have opposite signs. Yet \(a, b, c\) are of the same sign. Therefore two of \(a, b, c\) must be 0, and the other may be 1 or -1. Hence there are 6 possibilities for \((a, b, c)\), namely, \((\pm 1,0,0); (0, \pm 1,0)\) and \((0,0, \pm 1)\).