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Below example is equation of a plane example I would know how to determine the range of values for x, y and z respectively.

f = 5 x - 2 y + 7 z == 15;
ContourPlot3D[Evaluate[f],
 {x, -7, 7}, {y, -7, 7}, {z, -7, 10},
 Mesh -> None,
 ContourStyle -> Opacity[1],
 AspectRatio -> 1]

Thanks

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    $\begingroup$ I would know how to determine the range of values for x, y and z respectively It is not clear what you are asking. The range of values are the ones you gave in the command itself {x, -7, 7}, {y, -7, 7}, {z, -7, 10}. These are the values you decided to use. $\endgroup$ – Nasser Apr 11 at 6:45
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The question gave the shifted point-normal-form of a plane in 3D.

(x-p)n=0


n={5,-2,7}. d=np={5,-2,7}.{p1,p2,p2}=15.

A vector p is in general constructed from the normal

px={3,0,0}, py={0,-2/15,0} and pz={0,0,7/15}.

For graphical representation purposes, Mathematica has a built-in Hyperplane. That is in this question to be used in 3D. Just enter the normal vector to get a parallel plane through the origin.

Graphics3D[Hyperplane[{5, -2, 7}]]

The input for the shifted plane is

Graphics3D[Hyperplane[{5, -2, 7}, 15]]

A plot with the normal on the plane is

ill = Graphics3D[{Arrowheads[Medium], Thick, 
    Arrow[{{0, 0, 0}, {5, -2, 7}}]}, PlotRange -> 20, Axes -> True];
Show[ill, Graphics3D[Hyperplane[{5, -2, 7}, 15]]]

plane with normal

The range is 20 selected ad hoc. This corresponds to the setting Mathematica selects for the first input presented.

Mathematica can do more on planes: RegionMember:

RegionMember[Hyperplane[{5, -2, 7}, 15], {x, y, z}]

(x | y | z) \[Element] Reals && 5 x - 2 y + 7 z == 15

RegionDistance[Hyperplane[{5, -2, 7}, 15], {3, 0, 0}]
0

The point is on the plane.

There is no centroid of a plane. The plane is extended infinitely in two dimensions, directions. So range always arbitrary. Mathematica always shows a cube in three dimensions the standard view and viewport.

Have a look at the documentation of Hyperplane in Mathematica.

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