$16.2 .7^{\star}$ There is a two-digit number $x$, the unit digit and the tens digit are not the same. By swapping its unit digit and tens digit, we get another two-digit number $y$. It is also known that $x \cdot y$ equals a three-digit number $z$, and the hundreds digit and the unit digit of $z$ are the same. Find $x, y, z$.
Let $x=10a+b$, then $y=10b+a, xy=(10a+b)(10b+a)=100ab+10(a^2+b^2)+ab$.

Since $xy$ is a three-digit number, $ab<10$. Therefore, the unit digit and the hundred digit of $xy$ are both $ab$, and the ten digit is $a^2+b^2$. Since $a, b$ are different, non-zero, and $a^2+b^2 \leq 9$, we have $a=2, b=1$ or $a=1, b=2$. Thus, $x=21, y=12, z=252$ or $x=12, y=21, z=252$.