If the remainders of $x$, $y$, and $z$ divided by $3$ are all distinct, then we have $$x+y+z\equiv 0 + 1 +2 \equiv 0\pmod{3}$$

But if they are all distinct, then $(x-y)(y-z)(z-x)\not\equiv 0\pmod{3}$. Hence, this will contradict to the given condition. It follows that at least two of them are congruent modulo $3$. By symmetry, let's assume $x\equiv y\pmod{3}$.

Now we have $$\begin{array}{lll} &(x-y)(y-z)(z-x)\equiv 0 & \pmod{3} \\ \implies & x + y + z\equiv (x-y)(y-z)(z-x)\equiv 0 &\pmod{3} \\ \implies & z\equiv - (x+y)\equiv -2x \equiv x &\pmod{3} \\ \implies & x\equiv y\equiv z &\pmod{3}  \end{array} $$

$$\therefore (x-y)\equiv (y-z)\equiv(z-x)\equiv 0\pmod{3}\implies 27 \mid (x-y)(y-z)(z-x)$$

Hence, we find $27\mid (x+y+z)$ because $(x+y+z)= (x-y)(y-z)(z-x)$.