How to choose three vertices having maximum number of interconnecting edges?

You can use FindCycle and Subgraph:

g = Graph[UndirectedEdge @@@ edges];

Convert it to a simple graph since multiplicity doesn't necessary for this.

triangles = FindCycle[SimpleGraph[g], {3}, All];

subgraphs = Subgraph[g, #, AnnotationRules -> None] & /@ triangles;
VertexList[First[MaximalBy[subgraph, EdgeCount]]]

{5, 53, 61}

or

MaximalBy[triangles, 
 EdgeCount[Subgraph[g, #, AnnotationRules -> None]] &]

{{5 \[UndirectedEdge] 53, 53 \[UndirectedEdge] 61, 61 \[UndirectedEdge] 5}}

I would do it like this with IGraph/M:

Create a graph:

g = Graph[UndirectedEdge @@@ edges];

Merge parallel edges and record the multiplicities as edge weights:

wg = IGWeightedSimpleGraph[g];

Now we find the maximum total edge weight triangle:

MaximalBy[
  IGTriangles[g], 
  Total@IGEdgeProp[EdgeWeight]@IGWeightedSubgraph[wg, #] &
]

(* {{5, 61, 53}} *)

This will work in Mathematica 10.0 and later.

In Mathematica 12.0 or later, you can use Subgraph instead of IGWeightedSubgraph, as it does preserve weights.

A bit more complicated than @halmir's version, but I wanted to show how to work with edge weights. The main advantage of this answer is that IGTriangles is faster than FindCycle (because it is more specialized).

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Note: You may be thinking of this problem as finding maximum-weight 3-cliques. IGraph/M does have a function called IGWeightedCliques, but it works with vertex-weights, not edge-weights. I am noting this just to avoid confusion.

You can employ Szabolcs' package "IGraphM`" to find all triangles first (if there are any). Then finding the a triangle with maximal number of edges attached to it is straight-forward:

Needs["IGraphM`"]
G = Graph[Range[Max[edges]], UndirectedEdge @@@ edges];
triangles = IGTriangles[G];
i = Ordering[Total[Partition[VertexDegree[G][[Flatten[triangles]]], 3], {2}], -1][[1]];
triangles[[i]]