3. Find the positive integer solutions to the equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$.
3. Solution Since $x, y, z$ are positive integers and
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{6} \leqslant 1 .
$$

Assume without loss of generality that $1<x \leqslant y \leqslant z$, then $\frac{1}{x} \geqslant \frac{1}{y} \geqslant \frac{1}{z}$.
Thus, $\frac{1}{x}<\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{3}{x}$.
That is, $\frac{1}{x}<\frac{5}{6} \leqslant \frac{3}{x}$, solving this gives $\frac{5}{6}<x \leqslant \frac{18}{5}$, which means $x=2,3$.
When $x=2$, we can determine $y=4,5,6$.
When $x=3$, we can determine $y=3,4$.
In summary, we can determine $z=12,6,6,4$.

Therefore, when $1<x \leqslant y \leqslant z$, the solutions $(x, y, z)$ are
$$
\begin{array}{l}
(2,4,12),(2,6,6), \\
(3,3,6),(3,4,4)
\end{array}
$$

Four sets.
Since $x, y, z$ are symmetric in the equation, there are a total of 15 sets of solutions.