Minimum Possible Distance

As $\triangle ABC$ is equilateral, we have:

BD = $\frac{\sqrt3}{2}$ x $\sqrt3$ = $\frac{3}{2}$ m

Since P is equidistant from A and C, it will lie on the perpendicular bisector of AC.

Let PA = PC = a

Also, PB = PD + DB

$\implies$ PB = $\sqrt(a^{2} - (\frac{\sqrt3}{2})^{2}$ ) + $\frac{3}{2}$ = $\sqrt(a^{2} - \frac{3}{4})$ + $\frac{3}{2}$

PA + PB + PC = $\sqrt3$ ( $\sqrt3 + 2$ )

$\implies$ a + $\sqrt(a^{2} - \frac{3}{4})$ + $\frac{3}{2}$ + a = $\sqrt3$ ( $\sqrt3 + 2$ )

Solving for a, we get:

a = $\sqrt3$ or a = 2 + $\frac{5}{\sqrt3}$

From G (centroid as well as circumcenter of $\triangle ABC$ ), the distances from the vertices of the equilateral triangle will be the same. Also, G divides BD in the ratio 2:1.

Since PG should be the minimum possible, a must be equal to $\sqrt3$ .

PG = DG + PD

= [ $\frac{1}{3}$ x $\frac{3}{2}$ ] + $\frac{\sqrt(4a^{2} - 3)}{2}$

Putting a = $\sqrt3$ , we get:

= $\frac{1}{2}$ + $\frac{\sqrt(12 - 3)}{2}$ = 2 m