Example 1 Given $x, y, z \in Z_{+}, x \leqslant y \leqslant z$. Then the number of solutions to the equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}$ is ( ) groups.
(A) 8
(B) 9
(C) 10
(D) 11
(2016, Tsinghua University Independent Recruitment Examination)
Solve: From $0<x \leqslant y \leqslant z \Rightarrow \frac{1}{x} \geqslant \frac{1}{y} \geqslant \frac{1}{z}$ $\Rightarrow \frac{1}{x}<\frac{1}{2} \leqslant \frac{3}{x} \Rightarrow 2<x \leqslant 6$.
When $x=3$, the original equation becomes $\frac{1}{y}+\frac{1}{z}=\frac{1}{6}$.
Similarly, we get $6<y \leqslant 12$.
Since $x, y, z \in \mathbf{Z}_{+}$, verification shows that
$$
\begin{array}{l}
(x, y, z) \\
=(3,7,42),(3,8,24),(3,9,18), \\
\quad(3,10,15),(3,12,12) .
\end{array}
$$

Similarly, when $x=4,5,6$,
$$
\begin{array}{l}
(x, y, z) \\
=(4,5,20),(4,6,12),(4,8,8), \\
\quad(5,5,10),(6,6,6) .
\end{array}
$$

Therefore, the original equation has 10 solutions, choose (C).