Let the four colours be denoted by \(\{R,G,P,M\}\) (Red, Green, Purple, Magenta). The honeycomb is a regular hex grid. Two observations unlock the pattern:
1) Diagonal cycles.
Read the colours along any NE\(\leftrightarrow\)SW “diagonal” (constant axial coordinate in a hex grid). From several complete diagonals in the picture you can record sequences such as
\[
.s\; R,\,G,\,P,\,M,\,R,\,G,\,P,\,M\;.s
\]
on one diagonal, while the next diagonal is the same cycle but phase-shifted, e.g.
\[
.s\; G,\,P,\,M,\,R,\,G,\,P,\,M\;.s
\]
Hence, every such diagonal repeats the length-\(4\) cycle \((R\!\to G\!\to P\!\to M)\), only starting at a different place in the cycle.
2) Locate the phase for the diagonal that contains the three missing cells.
Look at the two filled hexes immediately to the left of the gap (on the same diagonal). Suppose they read \(\ldots, R, G\). By the cycle above, the next three colours on that diagonal must be \(P, M, R\) (continuing the repeating order).
Checking the neighbours around each of the three missing positions also confirms this: the colour chosen for each open cell must not duplicate any edge-adjacent hex. The sequence \(P, M, R\) satisfies all three local adjacency constraints visible around the gap; any alternative order causes at least one clash:
Options that place \(R\) first make the first new hex touch an existing \(R\) (edge adjacency conflict).
Options that place \(G\) in the middle clash with a neighbouring \(G\) above/below that position.
Options that place \(M\) last break the diagonal \(4\)-cycle (the next filled cell after the gap is \(G\), so the last in the gap must be \(R\) to keep \(\ldots,P,M,\underline{R},G,\ldots\)).
3) Match with the choices.
Among the four candidates, only (C) presents the required ordered triple on that diagonal (the same three colours in the same left\(\to\)right order) and simultaneously avoids all neighbour clashes. The other three choices either violate the diagonal cycle or create an immediate same-colour adjacency along one of the three insert positions.
Therefore the correct completion is \(\boxed{(C)}\).