Let XYZ be a three-digit number, where (X + Y + Z) is not a multiple of 3. Then (XYZ + YZX + ZXY) is not divisible by

• Let the three-digit number XYZ be represented as $100X + 10Y + Z$.

• The sum $S = XYZ + YZX + ZXY$ equals $(100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + Y)$.

• Simplifying the expression gives $S = 111X + 111Y + 111Z = 111(X + Y + Z)$.

• Since $111 = 3 \times 37$, the sum $S$ is always divisible by 3 and 37, regardless of the digits.

• The problem states that $(X + Y + Z)$ is not a multiple of 3, which implies $(X + Y + Z)$ is not divisible by 3.

• For $S$ to be divisible by 9, $111(X + Y + Z)$ must be divisible by 9. Since 111 is not divisible by 3 (sum of digits $1+1+1=3$, but $111/3=37$ which is not divisible by 3, so $111$ has only one factor of 3), the term $(X + Y + Z)$ must provide the remaining factor of 3 to make the product divisible by 9. However, since $(X + Y + Z)$ is not a multiple of 3, the product $111(X + Y + Z)$ cannot be divisible by 9.