Three, (50 points) Let $n$ be a positive integer. Find the number of positive integer triples $\langle a, b, c\rangle$ that satisfy the following conditions:
(i) $a b=n$;
(ii) $1 \leqslant c \leqslant b$;
(iii) The greatest common divisor of $a$, $b$, and $c$ is 1.
Let $(a, b, c)$ denote the greatest common divisor of $a, b, c$.
Let $S_{n}=\left\{\langle a, b, c\rangle \mid a, b, c \in \mathbf{Z}_{+}, a b=n\right.$,
$$
1 \leqslant c \leqslant b, (a, b, c)=1\} \text {. }
$$

Let the number of elements in $S_{n}$ be $f(n)\left(n \in \mathbf{Z}_{+}\right)$. Clearly, $f(1)=1$.
(1) If $n=p^{\alpha} (p$ is a prime, $\alpha \geqslant 1)$, let $\langle a, b, c\rangle \in S_{n}$.
If $b=1$, then $a=p^{\alpha}, c=1$;
If $b=p^{t}, 1 \leqslant t \leqslant \alpha-1$, then
$$
a=p^{\alpha-t}, (c, p)=1, 1 \leqslant c \leqslant b ;
$$

If $b=p^{\alpha}$, then $a=1, 1 \leqslant c \leqslant b$.
Thus, $f\left(p^{\alpha}\right)=1+\sum_{t=1}^{\alpha-1} \varphi\left(p^{t}\right)+p^{\alpha}$
$$
=p^{\alpha-1}+p^{\alpha} \text {, }
$$

where $\varphi(x)$ is the Euler's totient function.
(2) Next, we prove: if positive integers $m, n$ are coprime, then $f(m n)=f(m) f(n)$.
First, for each $\langle a, b, c\rangle \in S_{m n}$.
By $a b=m n$, let $b_{1}=(b, n), b_{2}=(b, m)$, then $\left(b_{1}, b_{2}\right)=1$;
Let $a_{1}=(a, n), a_{2}=(a, m)$, then $\left(a_{1}, a_{2}\right)=1$, and $a_{1} b_{1}=n, a_{2} b_{2}=m$.
By $1=(a, b, c)=\left(a_{1} a_{2}, b_{1} b_{2}, c\right)$ $=\left(\left(a_{1} a_{2}, b_{1} b_{2}\right), c\right)$
$$
=\left(\left(a_{1}, a_{2}\right)\left(b_{1}, b_{2}\right), c\right) \text {, }
$$

we know $\left(a_{1}, b_{1}, c\right)=1,\left(a_{2}, b_{2}, c\right)=1$.
Let $c_{i} \equiv c\left(\bmod b_{i}\right)\left(1 \leqslant c_{i} \leqslant b_{i}, i=1,2\right)$.
Then $\left(a_{1}, b_{1}, c_{1}\right)=1,\left(a_{2}, b_{2}, c_{2}\right)=1$
$$
\Rightarrow\left\langle a_{1}, b_{1}, c_{1}\right\rangle \in S_{n},\left\langle a_{2}, b_{2}, c_{2}\right\rangle \in S_{m} \text {. }
$$

Second, if $\left\langle a_{1}, b_{1}, c_{1}\right\rangle \in S_{n},\left\langle a_{2}, b_{2}, c_{2}\right\rangle \in S_{m}$, then let $a=a_{1} a_{2}, b=b_{1} b_{2}$.
Since $(m, n)=1$, thus, $\left(b_{1}, b_{2}\right)=1$.
By the Chinese Remainder Theorem, there exists a unique integer $c$ $(1 \leqslant c \leqslant b)$, satisfying
$$
\begin{array}{l}
\left\{\begin{array}{l}
c \equiv c_{1}\left(\bmod b_{1}\right), \\
c \equiv c_{2}\left(\bmod b_{2}\right) .
\end{array}\right. \\
\text { Clearly, }\left(a_{1}, b_{1}, c\right)=\left(a_{1}, b_{1}, c_{1}\right)=1 \text {, } \\
\left(a_{2}, b_{2}, c\right)=\left(a_{2}, b_{2}, c_{2}\right)=1 \text {. } \\
\text { Hence }(a, b, c)=((a, b), c) \\
=\left(\left(a_{1}, b_{1}\right)\left(a_{2}, b_{2}\right), c\right) \\
=\left(a_{1}, b_{1}, c\right)\left(a_{2}, b_{2}, c\right)=1 \\
\Rightarrow\langle a, b, c\rangle \in S_{m n} \\
\Rightarrow f(m n)=f(m) f(n) \text {. } \\
\end{array}
$$

Using (1) and (2), we know $f(n)=n \prod_{p i n}\left(1+\frac{1}{p}\right)$.