The distance from a fixed point $P$ on the plane to two vertices $A, B$ of an equilateral triangle $A B C$ are $A P=2 ; B P=3$. Determine the maximum value that the segment $P C$ can have.

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Let $A, B, C$ and $P$ be points on a plane such that $AB = BC = CA$, $AP = 2$, and $BP = 3$. Draw a ray $BM$ from point $B$ such that $\angle CBM = \angle ABP$, and mark a segment $BP' = PB$ on this ray. From the equality of angles: $\angle CBM = \angle ABP$, it follows that $\angle PBP' = \angle ABC = 60^\circ$, and therefore triangle $PBP'$ is equilateral (since $PB = BP'$). Hence, $PP' = PB = 3$. Furthermore, $\triangle PAB = \triangle P'CB$ (since $AB = BC$, $PB = P'B$, and $\angle ABP = \angle CBP'$).

Therefore, $P'C = PA = 2$. Thus, the broken line $PP'C$ has a length $PP' + P'C = 3 + 2 = 5$, and therefore the length of the segment $PC$ cannot be greater than 5. Choose point $A$ arbitrarily at a distance $AP = 2$ from the given point $P$.

Draw circles with radii $R_1 = 3$ and $R_2 = 5$ centered at point $P$ and rotate the second circle by an angle of $60^\circ$ relative to point $A$. Let $P'$ be the center of the new circle, $AP = AP' = 2$, and $\angle PAP' = 60^\circ$. Choose point $B$ as the intersection of the newly constructed circle with the circle of radius $R_1 = 3$ constructed earlier. Thus, the maximum possible distance from point $P$ to point $C$ is 5.

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