How to plot the given graph (irregular tri-hexagonal) with Mathematica?

Edit

Another effect by use ParametricPlot

a = 1;
{p1, p2, p3} = SSSTriangle[3/2 a, a, a][[1]];
e1 = p2 - p1;
e2 = p3 - p1;
fig = ParametricPlot[{u, v} . {e1, e2}, {u, 0, 10}, {v, 0, 10}, 
   MeshFunctions -> {#3 &, #4 &, #3 + #4 - 1 &}, 
   Mesh -> {Range[0, 20, 2]}, PlotPoints -> 80, Axes -> False, 
   Frame -> False, 
   MeshShading -> 
    ArrayReshape[ColorData[114, "ColorList"], {2, 2, 2}]];
pts = ParametricPlot[{u, v} . {e1, e2}, {u, 0, 10}, {v, 0, 10}, 
    MeshFunctions -> {#3 &, #4 &, #3 + #4 - 1 &}, 
    Mesh -> {Range[0, 20, 2]}, PlotStyle -> None, Axes -> False, 
    Frame -> False, BoundaryStyle -> None, PlotPoints -> 100] // 
   Graphics`Mesh`FindIntersections;
Show[fig, Graphics[{PointSize[Large], Black, Point[pts]}]]

Updated

a = 1;
SSSTriangle[3/2 a, a, a];
{p1, p2, p3} = %[[1]];
lines1 = Table[
   TranslationTransform[i*2 (p2 - p1)][
    InfiniteLine[{p1, p3}]], {i, -5, 5}];
lines2 = Table[
   TranslationTransform[i*2 (p3 - p1)][
    InfiniteLine[{p1, p2}]], {i, -5, 5}];
lines3 = Table[
   TranslationTransform[i*2 (p2 - p1)][
    InfiniteLine[{p2, p3}]], {i, -5, 5}];
points1 = Outer[RegionIntersection[{#1, #2}] &, lines1, lines2];
points2 = Outer[RegionIntersection[{#1, #2}] &, lines2, lines3];
points3 = Outer[RegionIntersection[{#1, #2}] &, lines3, lines1];
Graphics[{lines1, lines2, lines3, Red, PointSize[Large], points1, 
  points2, points3}, PlotRange -> 5]

Original

a = 1;
SSSTriangle[3/2 a, a, a];
{p1, p2, p3} = %[[1]]
Graphics[{Table[
   TranslationTransform[i*2 (p2 - p1)][
    InfiniteLine[{p1, p3}]], {i, -5, 5}], 
  Table[TranslationTransform[i*2 (p3 - p1)][
    InfiniteLine[{p1, p2}]], {i, -5, 5}], 
  Table[TranslationTransform[i*2 (p2 - p1)][
    InfiniteLine[{p2, p3}]], {i, -5, 5}]}, PlotRange -> 5]

Update: We can generalize the method in the original answer to make the side length a parameter:

ClearAll[angleList, anglePath, hexaGon, hexTile]

angleList[α_] := TriangleMeasurement[SSSTriangle[α, 1, 1], {"InteriorAngle", All}] 

angleList[α]

{ArcCos[1 - α^2/2], ArcCos[Sqrt[α^2]/2],  ArcCos[Sqrt[α^2]/2]}  

anglePath[α_] := FullSimplify @ AnglePath[Prepend[{1, 0}] @
     PadRight[Thread[{{1, α, 1}, angleList[α]}], 5, "Periodic"]]

hexaGon[α_] := {Black, Line[anglePath[α]], Red, PointSize @ Large, Point @ anglePath[α]}

Examples:

Row[{Graphics[hexaGon[1], ImageSize -> 1 -> 100, PlotLabel -> Style["α = 1", 16]], 
  Graphics[hexaGon[3/2], ImageSize -> 1 -> 100, PlotLabel -> Style["α = 3/2", 16]], 
  Graphics[hexaGon[5/4], ImageSize -> 1 -> 100, PlotLabel -> Style["α = 5/4", 16]], 
  Graphics[hexaGon[2/3], ImageSize -> 1 -> 100, PlotLabel -> Style["α = 2/3", 16]]},
     Spacer[20]]

hexTile[α_, nc_, nr_, opts : OptionsPattern[]] := 
 Module[{tr = Subtract @@ anglePath[α][[{1, 4}]]}, 
  Graphics[Table[Translate[hexaGon[α], 
     {2 i - j First[tr], -j Last[tr]}], {i, nc}, {j, 0, nr - 1}],
   opts, ImageSize -> Large]]

Examples:

hexTile[1, 7, 5]

hexTile[3/2, 7, 5]

hexTile[5/4, 7, 5]

hexTile[2/3, 7, 5]

Original answer:

angles = TriangleMeasurement[SSSTriangle[3/2, 1, 1], {"InteriorAngle", All}] 

{ArcCos[-(1/8)], ArcCos[3/4], ArcCos[3/4]}  

We can use angles and desired lengths (1, 3/2 and 1) with AnglePath to get the coordinates of desired hexagon primitive:

anglepath = FullSimplify @ AnglePath[Prepend[{1, 0}] @
        PadRight[Thread[{{1, 3/2, 1}, angles}], 5, "Periodic"]];

hex = {Line @ anglepath, Red, PointSize @ Large, Point @ anglepath};

Graphics[hex]

We get the desired picture using translations of hex:

Graphics[Table[Translate[hex, {2 i - j/4, j  3 Sqrt[7]/4}], {i, 5}, {j, 0, 5}]] 

Graphics[Table[Translate[hex, {2 i - j/4, j  3 Sqrt[7]/4}], {i, 10}, {j, 0, 4}], 
 ImageSize -> Large]