Let the point P be the vertex of the parabola y = x^2 - 6x + 12 ....

The equation of the parabola is $y = x^2 - 6x + 12$, which can be rewritten as $y - 3 = (x - 3)^2$.

The vertex of the parabola is $P(3, 3)$.

The equation of the circle is $x^2 + y^2 - 2x - 4y + 3 = 0$, which can be rewritten as $(x - 1)^2 + (y - 2)^2 = 2$.

The center of the circle is $C(1, 2)$ and its radius is $r = \sqrt{2}$.

The distance between the point $P$ and the center $C$ is:

$PC = \sqrt{(3 - 1)^2 + (3 - 2)^2} = \sqrt{4 + 1} = \sqrt{5}$

Since $PC = \sqrt{5} > r = \sqrt{2}$, the point $P$ lies outside the circle.

Let the line through $P$ intersect the circle at $R$ and $S$, and let $M$ be the midpoint of the chord $RS$. The perpendicular distance from the center $C$ to the chord is $CM$.

Using the right-angled triangle $\triangle PCM$, the distance from $P$ to the midpoint $M$ is $PM = \sqrt{PC^2 - CM^2}$.

The lengths of the segments are $PR = PM - RM$ and $PS = PM + RM$.

Therefore, the sum of the distances is:

$PR + PS = (PM - RM) + (PM + RM) = 2PM = 2\sqrt{PC^2 - CM^2}$

To maximize $(PR + PS)^2$, we need to maximize $PC^2 - CM^2$. This occurs when $CM$ is minimized, which means $CM = 0$. This happens when the line passes through the center $C$ of the circle.

Substituting $CM = 0$, we get the maximum sum:

$PR + PS = 2PC = 2\sqrt{5}$

Thus, the maximum value of $(PR + PS)^2$ is:

$(2\sqrt{5})^2 = 4 \times 5 = 20$

Answer: $20$