Three. (25 points) Given that $a$, $b$, and $c$ are positive integers, and $a < b < c$. If
$$
a b c \mid (a b - 1)(b c - 1)(c a - 1),
$$

then, can three segments of lengths $\sqrt{a}$, $\sqrt{b}$, and $\sqrt{c}$ form a triangle? If so, find the area of the triangle; if not, explain the reason.
Three, notice that
$(a b-1)(b c-1)(c a-1)$
$=a b c(a b c-a-b-c)+a b+b c+c a-1$.
From $a b c \mid(a b-1)(b c-1)(c a-1)$, we get
$a b c \mid(a b+b c+c a-1)$,
which means $\frac{a b+b c+c a-1}{a b c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a b c}$
is an integer.
Since $a b+b c+c a-1>0$, we know that
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a b c}
$$

is a positive integer.
Therefore, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>1$.
If $a=1$, then $\frac{1}{b}+\frac{1}{c}-\frac{1}{b c}$ is a natural number.
Since $\frac{1}{b}+\frac{1}{c} \leqslant \frac{1}{2}+\frac{1}{3}<1$, it follows that
$$
\frac{1}{b}+\frac{1}{c}-\frac{1}{b c}=0 \Rightarrow b+c=1,
$$

which is a contradiction.
If $a \geqslant 3$, then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{3}{a} \leqslant 1$, which is a contradiction. Therefore, $a=2$.

If $b \geqslant 4$, then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1$, which is a contradiction. Therefore, $b=3$.

If $c \geqslant 6$, then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$, which is a contradiction. Therefore, $c=4$ or 5.
Upon inspection, only $a=2, b=3, c=5$ satisfy the conditions.
In this case, $\sqrt{a}=\sqrt{2}, \sqrt{b}=\sqrt{3}, \sqrt{c}=\sqrt{5}$.
Since $(\sqrt{2})^{2}+(\sqrt{3})^{2}=(\sqrt{5})^{2}$, the three line segments with lengths $\sqrt{a}, \sqrt{b}, \sqrt{c}$ can form a right triangle, with an area of
$$
S=\frac{1}{2} \times \sqrt{2} \times \sqrt{3}=\frac{\sqrt{6}}{2} .
$$