= Solution
Three PO$_4$ tetrahedra share corners in a ring. Each <phosphorus> has two ring-bridging <oxygens> and two terminal <oxygens>, so
$$
\boxed{\left[\mathrm{{-}P(O)(O^-){-}O{-}P(O)(O^-){-}O{-}P(O)(O^-){-}O{-}}\right]^{3-}_{\rm cyclic}}
$$
is one conventional Lewis representation of <trimetaphosphate>. Its three P and nine O atoms consist of three bridging and six terminal O atoms. Within each phosphate unit the two terminal bonds have delocalized character; drawing one P=O and one P–O$^-$ is not a claim that every free <anion> has permanently different terminal <oxygens>.
\b[A puckered six-membered P–O ring is expected.] The tetrahedral <phosphorus> geometry and P–O–P torsional preferences allow folding without losing a required continuous perpendicular-p orbital system. This differs from the planar trigonal-boron arrangement. Tetrahedral P alone is not a geometrical proof of nonplanarity, since its two ring bonds could lie in a plane; the argument is that puckering is energetically accessible and relieves conformational strain without sacrificing a strong planar pi-conjugation benefit. The precise chair-like or boat-like distortion depends on environment and counterions.
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