Seven, (25 points) Find
$$
[x, y, z]=(x, y)+(y, z)+(z, x)
$$

satisfying $x \leqslant y \leqslant z,(x, y, z)=1$ for all positive integer solutions, where $[m, n]$ and $(m, n)$ represent the least common multiple and greatest common divisor of positive integers $m, n$, respectively.
Seven, because $(x, y, z)=1$, so,
$\frac{(x, y)}{[x, y, z]}=\frac{(x, y)}{[[x, y], z]}$
$\leqslant \frac{(x, y)}{[(x, y), z]}=\frac{(x, y)}{(x, y) z}=\frac{1}{z}$.
Similarly, $\frac{(x, z)}{[x, y, z]} \leqslant \frac{1}{x}, \frac{(z, x)}{[x, y, z]} \leqslant \frac{1}{y}$.
By the given condition, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geqslant 1$.
Also, $x \leqslant y \leqslant z$, then $\frac{1}{x} \geqslant \frac{1}{y} \geqslant \frac{1}{z}$. Hence $\frac{1}{x} \geqslant \frac{1}{3}$.
If $\frac{1}{x}=\frac{1}{3}$, then $x=y=z=3$, which contradicts $(x, y, z)=1$.
Therefore, $\frac{1}{x}>\frac{1}{3} \Rightarrow x<3 \Rightarrow x=1,2$.
(1) If $x=1$, then $[y, z]=(y, z)+2$.
Thus, $(y, z)$ |2. So, $(y, z)=1,2$.
When $(y, z)=1$, the solution is $y=1, z=3$.
When $(y, z)=2$, the solution is $y=2, z=4$.
(2) If $x=2$, then $\frac{1}{y}+\frac{1}{z} \geqslant \frac{1}{2}$, so $\frac{1}{y} \geqslant \frac{1}{4}$.
Thus, $2 \leqslant y \leqslant 4$.
When $y=2$, $[2, z]=2+2(2, z)$, no solution.
When $y=3$, $3 \leqslant z \leqslant 6$.
Upon verification, $z=6$.
When $y=4$, only $z=4$, which contradicts $(x, y, z)=1$.
In summary, there are 3 solutions:
$$
(x, y, z)=(1,1,3),(1,2,4),(2,3,6) .
$$