Three, (16 points) Given the inverse proportion function $y=\frac{6}{x}$ and the linear function $y=k x-5$ both pass through the point $P(m, 3)$.
(1) Find the value of $k$;
(2) Draw a line parallel to the $y$-axis, intersecting the graph of the linear function at point $A$ and the graph of the inverse proportion function at point $B$. Let the x-coordinate of points $A$ and $B$ be $a$. Find the value of $a$ when $P A=P B$?
(1) Substituting the coordinates of point $P(m, 3)$ into the two functions, we get
$$
\left\{\begin{array}{l}
3=\frac{6}{m}, \\
3=k m-5 .
\end{array}\right.
$$

Solving, we get $m=2, k=4$.
(2) As shown in Figure 6, draw $P E \perp A B$. To make $P A=P B$, we only need $B E=E A$.
Since the x-coordinates of $A$ and $B$ are $a$, we have
$$
\begin{array}{l}
A(a, 4 a-5), B\left(a, \frac{6}{a}\right), E(a, 3) . \\
\quad \text { Substituting } B E=E A \text {, we get } \frac{6}{a}-3=3-(4 a-5) .
\end{array}
$$

Substituting $B E=E A$, we get $\frac{6}{a}-3=3-(4 a-5)$.
Simplifying, we get $(a-2)(4 a-3)=0$.
Solving, we get $a_{1}=2, a_{2}=\frac{3}{4}$.
Since the intersection point $Q\left(-\frac{3}{4},-8\right)$ of the line $y=4 x-5$ and the inverse proportion function $y=\frac{6}{x}$ also satisfies the given conditions, we have $a_{3}=-\frac{3}{4}$.
Therefore, $a=2$ or $a= \pm \frac{3}{4}$.