Compounds having planar symmetry is
A) \[{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\]
B) \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\]
C) \[{\rm{HN}}{{\rm{O}}_{\rm{3}}}\]
D) \[{\rm{CC}}{{\rm{l}}_{\rm{4}}}\]

Hint: A trigonal planar symmetry is the molecular geometry acquired by the \[s{p^2}\] hybridized molecules. An \[s{p^2}\]hybridized molecule has three electron groups surrounding it. Some \[s{p^2}\]hybridized compounds are \[{\rm{B}}{{\rm{F}}_{\rm{3}}},{\rm{BC}}{{\rm{l}}_{\rm{3}}}\] .

Complete step by step solution:Let’s discuss the VSEPR theory in detail. This theory helps to determine the shapes of the molecules considering the lone pairs and bond pairs that surround the centrally placed atom. According to this theory, the order of repulsion between bond pairs and lone pairs is lone pair-lone pair>lone pair-bond pair>bond pair-bond pair.
Let’s discuss all the options one by one.
Option A is \[{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\]. This central atom of the molecule, that is, Sulphur, is surrounded by four electron groups (O, O, OH, OH). So, it is \[s{p^3}\] hybridized. And the molecular geometry is tetrahedral.
Option B is \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\]. The electron groups surrounding the O atom are four (Two oxygen atoms and two lone pairs). So, it is \[s{p^3}\] hybridized. And geometry is tetrahedral and the shape is bent.
Option C is\[{\rm{HN}}{{\rm{O}}_{\rm{3}}}\]. The electron groups surrounding the N atom are three oxygen atoms. Therefore, it is \[s{p^2}\] hybridized. Therefore, the molecular geometry is trigonal planar.
Option D is \[{\rm{CC}}{{\rm{l}}_{\rm{4}}}\]. Four chlorine atoms surround the carbon atom. Therefore, it is \[s{p^3}\] hybridized. Therefore, the molecular geometry is tetrahedral.

Therefore, option C is right.

Note: If the molecular geometry is trigonal planar and there is no presence lone pair, then the molecular shape would also be trigonal planar. But, if two bond pair and a lone pair are present, then the molecular shape would be bent.