How can I express the integral path of a curve integral?

Question

A second sort of curve integral:

$\oint_{\Gamma}\left(y^{2}-z^{2}\right) d x+\left(z^{2}-x^{2}\right) d y+\left(x^{2}-y^{2}\right) d z$

The integral path $\Gamma$ is the cut-off line obtained by cutting the surface of the cube($\{(x, y, z) \mid 0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1,0 \leqslant z \leqslant 1\}$) with a plane ($x+y+z=\frac{3}{2}$). If you look from the forward direction of the ox axis, take the counterclockwise direction.

You can find the integration path in this 3D graph. It is the boundary of the hexagon, and the direction is counterclockwise:

Clear["Global`*"];
ContourPlot3D[{x + y + z == 3/2, x == 0, x == 1, y == 0, z == 0,
    y == 1, z == 1}, {x, #, #2}, {y, #3, #4}, {z, #5, #6}, Mesh -> 1,
   ImageSize -> 500, ViewPoint -> {5, 1, 2},
   RegionFunction ->
    Function[{x, y,
      z}, \!\(TraditionalForm\`\(-0.1\) ⩽
          x ⩽ 1.1 && \(-0.1\) ⩽
          y ⩽ 1.1 && \(-0.1\) ⩽
          z ⩽ 1.1\) && x + y + z <= 1.6],
   ContourStyle -> (Directive @@@ {{Red, [email protected]}, {Green,
        [email protected]}, {Blue, [email protected]}, {Yellow, [email protected]}, {Cyan,
        [email protected]}, {Gray, [email protected]}, {Pink, [email protected]}}),
   Lighting -> "Neutral", Axes -> True, AspectRatio -> 1,
   AxesOrigin -> {0, 0, 0}, AxesLabel -> {"x", "y", "z"},
   PlotPoints -> 50] & @@@ {{-1, 2, -1, 2, -1, 2}}

I tried to write the code of integral path, but RegionBoundary returns the original plane region as the boundary.

Clear["Global`*"];
plane = ImplicitRegion[
  x + y + z == 1.5 && 0 <= x <= 1 && 0 <= y <= 1 && 0 <= z <= 1, {x,
   y, z}];
line = RegionBoundary[plane]

ImplicitRegion[x+y+z\[Equal]1.5&&0<=x<=1&&0<=y<=1&&0<=z<=1,{x,y,z}]

Answer 1

Here's a possible work-around:

poly = Cube[1/2 {1, 1, 1}, 1] // RegionBoundary

lines = Line@poly[[1]][[#]] & /@ poly[[2]]

pts={x, y, z} /. 
    Solve @@ RegionIntersection[#, ImplicitRegion[x + y + z == 3/2, {x, y, z}]] &@
  RegionConvert[#, "Implicit"] & /@ lines // Flatten[#, 1] & // DeleteDuplicates
(* {{1/2, 1, 0}, {1, 1/2, 0}, {1, 0, 1/2}, 
    {1/2, 0, 1}, {0, 1/2, 1}, {0, 1, 1/2}} *)

At this point, the order of pts happens to be desired:

Arrow@Append[pts, pts[[1]]] // Graphics3D

But using FindCurvePath should make the solution a bit more general:

ptsordered = pts[[First@FindCurvePath@pts]];

ListLinePlot3D@ptsordered /. Line -> Arrow

Answer 2

poly = RegionIntersection[Hyperplane[{1, 1, 1}, {0, 0, 3/2}], Cuboid[]]

Polygon[{{1, 0, 1/2}, {1/2, 0, 1}, {0, 1/2, 1}, {0, 1, 1/2}, {1/2, 1, 0}, {1, 1/2, 0}}]

or

reg=ImplicitRegion[x+y+z==3/2&&0<=x<=1&&0<=y<=1&&0<=z<=1,{x,y,z}];
poly=MeshPrimitives[Region`Mesh`MergeCells[DiscretizeRegion[reg]],2][[1]]

Polygon[{{1., 0.5, 0.}, {1., 0., 0.5}, {0.49999999999999994, 0., 1.}, {0., 0.49999999999999994, 1.}, {0., 1., 0.5}, {0.49999999999999994, 1., 0.}}]

paraEq={Abs[t-1]+3/2 Abs[t-1/2]+3/2 Abs[t-1/3]-3/2 Abs[t-5/6]-Abs[t]/2,3/2 Abs[t-1]-3/2 Abs[t-1/2]-3/2 Abs[t-2/3]+3/2 Abs[t-1/6],-Abs[t-1]+2 Abs[t]-3/2 Abs[t-1/3]+3/2 Abs[t-2/3]-3/2 Abs[t-1/6]+3/2 Abs[t-5/6]};

paraEq2={Sqrt[3]/2  Cos[θ] Sec[1/6 ArcCos[-Cos[6 θ]]],Sqrt[3]/2  Sec[1/6 ArcCos[-Cos[6 θ]]] Sin[θ],1}.{{0,-(1/2),1/2},{-(1/Sqrt[3]),1/(2 Sqrt[3]),1/(2 Sqrt[3])},{1/2,1/2,1/2}};

Show[Graphics3D[poly],
ParametricPlot3D[paraEq,{t,0,1}],
ParametricPlot3D[paraEq2,{θ,0,2Pi}]]

rule=Thread[{x,y,z}->paraEq];
Integrate[Hold[D[{x,y,z},t].{y^2-z^2,z^2-x^2,x^2-y^2}]/.rule/.Abs->RealAbs//ReleaseHold,{t,1,0}]

$-\frac{9}{2}$