2. All positive rational triples $(x, y, z) (x \leqslant y \leqslant z)$ that make $x+y+z, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ and $x y z$ integers are $\qquad$ .
$$
\text { 2. }(x, y, z)=(1,1,1),(1,2,2),(2,3, }
$$
$6),(2,4,4),(3,3,3) 5$ groups.
$$
\text { Let } \begin{aligned}
f(u)= & (u-x)(u-y)(u-z) \\
= & u^{3}-(x+y+z) u^{2}+(x y+y z \\
& +z x) u-x y z .
\end{aligned}
$$
$f(u)$ is a polynomial with integer coefficients. Given this, the roots must all be integers, so $x, y, z$ must all be integers. By $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ being a positive integer, when $x=1, y \geqslant 3$, $\frac{1}{y}+\frac{1}{z} \leqslant \frac{2}{3}$ has no solution. When $x=2, y \geqslant 5$, $\frac{1}{y}+\frac{1}{z} \leqslant \frac{2}{5}$ also has no solution. When $x \geqslant 3, y \geqslant 4$, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<1$ has no solution. The only integer solutions are the 5 groups mentioned above.