We can write $x = 10a + b$, $y = 10b + c$, $z = 10c + a$ with positive integers $a, b, c$ less than $10$. Let $d$ be the greatest common divisor of $x$, $y$ and $z$. $11$ cannot divide $d$, since otherwise it contradicts the fact that $x$, $y$ and $z$ are distinct.

$x + y + z = (10a + b) + (10b + c) + (10c + a) = 11(a + b + c)$ is divisible by $d$. Since $d$ is not divisible by $11$, $a+b+c$ is divisible by $d$. Thus $d \le a+b+c \le 27$.

Also, $100x - 10y + z = 1000a + 100b - 100b - 10c + 10c + a = 1001a$ is divisible by $d$. Similarly, $1001b$ and $1001c$ are divisible by $d$. So $1001k$ and hence $91k$ is divisible by $d$, where $k$ is the greatest common divisor of $a$, $b$ and $c$.

Since $d \le 27$ and $k \le 4$ (otherwise $x = y = z$), $d$ is one of $1, 2, 3, 4, 7, 13, 14, 21$ or $26$.

But, $d$ cannot be $21$, because a 2-digit multiple of $21$ must be $21, 42, 63$ or $84$, and we can't find $x$, $y$, $z$ among them with the required conditions. Similarly, $d$ cannot be $26$.

On the other hand, since $(x, y, z) = (32, 21, 13), (64, 42, 26), (96, 63, 39), (88, 84, 48), (42, 21, 14), (65, 52, 26)$ and $(84, 42, 28)$ give examples for $d = 1, 2, 3, 4, 7, 13$ and $14$, there are exactly $7$ such $d$'s.