Three. (25 points) Given that $P(2,3)$ is a point on the graph of the inverse proportion function $y=\frac{k}{x}$.
(1) Find the equation of the line passing through point $P$ and having only one common point with the hyperbola $y=\frac{k}{x}$.
(2) $Q$ is a moving point on the branch of the hyperbola $y=\frac{k}{x}$ in the third quadrant. Draw a line through point $Q$ such that it has only one common point with the hyperbola $y=\frac{k}{x}$, and intersects the $x$-axis and $y$-axis at points $C$ and $D$, respectively. Let the line found in (1) intersect the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Determine the positional relationship between $A D$ and $B C$.
(3) Based on (2), analyze the shape of quadrilateral $A B C D$ when its area is minimized.
(1) From the given conditions, it is easy to obtain the equation of the hyperbola as $y=\frac{6}{x}$. Clearly, the lines $x=2$ and $y=3$ meet the requirements. Let the equation of the third line be $y=a(x-2)+3$.
Then $\left\{\begin{array}{l}y=a(x-2)+3, \\ y=\frac{6}{x}\end{array}\right.$, which has only one solution.
Eliminating and rearranging gives
$$
a x^{2}+(3-2 a) x-6=0 \text {. }
$$

Thus, $\Delta=(2 a+3)^{2}=0$. Solving for $a$ gives $a=-\frac{3}{2}$.
Therefore, the equation of the third line is $y=-\frac{3}{2} x+6$.
(2) From (1), we know $A(4,0)$ and $B(0,6)$.
Let the equation of line $CD$ be $y=m x+n$.
Then $\left\{\begin{array}{l}y=m x+n, \\ y=\frac{6}{x}\end{array}\right.$ has only one solution.
Eliminating and rearranging gives
$$
m x^{2}+n x-6=0 \text {. }
$$

Thus, $\Delta=n^{2}+24 m=0$.
Therefore, $O C \cdot O D=\frac{n}{m}(-n)=24=O A \cdot O B$, which means
$$
\frac{O A}{O C}=\frac{O D}{O B} \text {. }
$$

Hence, $A D / / B C$.
(3) Let $O C=t$, then
$S_{\text {quadrilateral } H C D}=3 t+\frac{48}{t}+24=3\left(\sqrt{t}-\frac{4}{\sqrt{t}}\right)^{2}+48$.
When this area is minimized, $t$ is clearly 4, then
$$
O A=O C=4, O B=O D=6 \text {. }
$$

Since $A C \perp B D$, the quadrilateral $A B C D$ is a rhombus.