How to return the coordinates of the respective 3 vertexes of a DelaunayMesh's face?

Finding the correct triangle is a very delicate business. See for example:

pts1 = BlockRandom[SeedRandom[1]; RandomReal[1, {30, 2}]];
R = DelaunayMesh[pts1];
redPt = {0.68, 0.75};
idx = Nearest[MeshCoordinates[R] -> Automatic, redPt, 3];
r = Max@Nearest[MeshCoordinates[R] -> "Distance", redPt, 3];
i = Position[Through[(RegionMember /@ MeshPrimitives[R, 2])[redPt]], 
    True][[1, 1]];
Show[
 HighlightMesh[R, Join[Thread[{0, idx}], {{2, i}}]],
 Graphics[{
   Red, Point@redPt,
   Circle[redPt, r]
   }],
 PlotRange -> All
 ]

Fortunately, there seems to be a built-in but undocumented function for this task:

j = Region`Mesh`MeshNearestCellIndex[R, redPt];
MeshPrimitives[R, j]
Show[
 HighlightMesh[R, j],
 Graphics[{Red, Point[redPt]}]
 ]

Polygon[{{0.925275, 0.578056}, {0.767697, 0.973336}, {0.544772, 0.562659}}]

Update: As noted by Henrik and Rahul the original answer is not the correct approach. An alternative method that gives the correct polygon containing redPt is based on using Graphics`PolygonUtils`InPolygonQ:

ClearAll[f]
f[r_, p_] :=  Pick[#, Graphics`PolygonUtils`InPolygonQ[#, p] & /@ #] &@
   MeshPrimitives[r, 2];

Show[plot1, 
 Graphics[{PointSize[Large], Point@redPt, Opacity[.5, Red],  f[plot1, redPt]}]] 

To get the coordinates, use

f[plot1, redPt][[1, 1]]

{{0.67033, 0.84245}, {0.620283, 0.648944}, {0.829905, 0.700287}}

Using Henrik's example (R):

Show[R, Graphics[{PointSize[Large], Point@redPt, Opacity[.5, Red],  f[R, redPt]}]]

Original (incorrect) answer:

Nearest[MeshCoordinates[plot1], redPts, 3]