One, (20 points) The vertex of the quadratic function $y=x^{2}-4 x+3$ is $P$. A line $A B$ passing through the point $\left(\frac{3}{2},-\frac{7}{4}\right)$ is tangent to the parabola, and the line $A B$ passes through the first, third, and fourth quadrants, intersecting the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Find the distance from point $P$ to the line $A B$.
1. As shown in Figure 7, let the equation of the line $AB$ be $y = kx + b$.
   Substituting $x = \frac{3}{2}$ and $y = -\frac{7}{4}$, we get
   $$
   b = -\frac{7}{4} - \frac{3}{2}k.
   $$
   Therefore, $y = kx - \frac{3}{2}k - \frac{7}{4}$.
   From the system of equations
   $$
   \left\{\begin{array}{l}
   y = kx - \frac{3}{2}k - \frac{7}{4}, \\
   y = x^2 - 4x + 3,
   \end{array}\right.
   $$
   eliminating $y$ gives
   $$
   x^2 - (4 + k)x + \left(\frac{3}{2}k + \frac{19}{4}\right) = 0.
   $$

   Since the line $AB$ is tangent to the parabola, we have
   $$
   \Delta = (4 + k)^2 - 4\left(\frac{3}{2}k + \frac{19}{4}\right) = 0.
   $$

   Solving this, we get $k = 1$ or $k = -3$.
   Given that $k > 0$, we have $k = 1$.
   Therefore, the equation of the line $AB$ is $y = x - \frac{13}{4}$. It is easy to see that $A\left(\frac{13}{4}, 0\right)$ and $B\left(0, -\frac{13}{4}\right)$, so $OA = \frac{13}{4}$ and $OB = \frac{13}{4}$.
   In the right triangle $\triangle AOB$, $AB = \sqrt{OA^2 + OB^2} = \frac{13\sqrt{2}}{4}$.
   Let the distance from point $P$ to the line $AB$ be $d$. Connecting $PO$, $PA$, and $PB$, we have
   $$
   \frac{1}{2}OA \cdot 1 + \frac{1}{2}OB \cdot 2 + \frac{1}{2}AB \cdot d = \frac{1}{2}OA \cdot OB.
   $$

   Considering $P(2, -1)$, we get
   $$
   d = \frac{OA \cdot OB - OA - 2OB}{AB} = \frac{\sqrt{2}}{8}.
   $$
   Therefore, the distance from point $P$ to the line $AB$ is $\frac{\sqrt{2}}{8}$.