Three. (20 points) Given a parabola $y^{2}=2 p x(p>0)$ with a chord $A B$ of length $l$. Find the shortest distance from the midpoint $M$ of $A B$ to the $y$-axis, and determine the coordinates of point $M$ at this time.

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Three, from the parabolic equation $y^{2}=2 p x$, we can set the coordinates of $A$ and $B$ as $A\left(\frac{a^{2}}{2 p}, a\right), B\left(\frac{b^{2}}{2 p}, b\right)$. Then the coordinates of the midpoint $M$ are
$$
\left\{\begin{array}{l}
x=\frac{a^{2}+b^{2}}{4 p} . \\
y=\frac{a+b}{2} .
\end{array}\right.
$$

We have $(a+b)^{2}=4 y^{2}$,
$$
(a-b)^{2}=8 p x-4 y^{2} \text {. }
$$

Also, given $|A B|=l$, we have
$$
\begin{aligned}
l^{2} & =\left(\frac{a^{2}-b^{2}}{2 p}\right)^{2}+(a-b)^{2} \\
& =(a-b)^{2} \cdot \frac{(a+b)^{2}+4 p^{2}}{4 p^{2}} .
\end{aligned}
$$

Substituting equations (1) and (2) into this, we get
$$
4\left(2 p x-y^{2}\right)\left(y^{2}+p^{2}\right)-p^{2} l^{2}=0 \text {. }
$$
(1) When $l \geqslant 2 p$, from equation (3) we have
$$
\begin{array}{l}
x=\frac{1}{2 p}\left[\left(y^{2}+p^{2}\right)+\frac{p^{2} l^{2}}{4\left(y^{2}+p^{2}\right)}\right]-\frac{p}{2} \\
\geqslant \frac{1}{p} \sqrt{\frac{p^{2} l^{2}}{4}}-\frac{p}{2}=\frac{l-p}{2} .
\end{array}
$$

Equality holds if and only if
$$
y^{2}+p^{2}=\frac{p^{2} l^{2}}{4\left(y^{2}+p^{2}\right)} \Rightarrow y=\sqrt{\frac{p l}{2}-p^{2}} .
$$

Thus, the shortest distance from point $M$ to the $y$-axis is $x_{\operatorname{min}}=\frac{l-p}{2}$. At this point, the coordinates of $M$ are $\left(\frac{l-p}{2}, \sqrt{\frac{p l}{2}-p^{2}}\right)$.

Explanation: The geometric meaning of this problem is as shown in Figure 4. Let $F$ be the focus of the parabola, and $CD$ be the directrix, then
$$
\begin{aligned}
x_{M} & =M N-\frac{p}{2} \\
& =\frac{A C+B P}{2}-\frac{p}{2} \\
& =\frac{F A+F B}{2}-\frac{p}{2} \\
& \geqslant \frac{A B}{2}-\frac{p}{2}=\frac{l-p}{2} .
\end{aligned}
$$

When $A B$ passes through the focus $F$, $x_{\min }=\frac{l-p}{2}$. But when $l < 2p$,
it is evident that the function (1) is increasing with respect to the variable $y^{2}$. When $y=0$, the minimum value is $x_{\text {min }}=\frac{l^{2}}{8 p}$, and the coordinates of point $M$ are $\left(\frac{l^{2}}{8 p}, 0\right)$.
In summary, $x_{\text {min }}=\left\{\begin{array}{ll}\frac{l^{2}}{8 p}, & 0<l<2 p \text {, } \\ \frac{l-p}{2}, & l \geqslant 2 p \text {, }\end{array}\right.$