Let's suppose that you are asking how to solve the system of linear equations M . XYZ = V where M is your given Matrix, V is your given Vector, and XYZ is a Vector with entries as three variables. You can use Maple's LinearAlgebra package as solve that system (without even needing to specify XYZ, or you can form the three explicit equations and use Maple's solve command on those.
M := Matrix([ [ 863138, -332862, -530302 ],
[ 1726276, -665724, -530302*m ],
[ -863138, -998586, -1060604 ]]):
V := Vector([ 8, -8, -4*m ]):
First way,
with(LinearAlgebra):
sol1 := LinearAlgebra:-LinearSolve(M,V);
[ 2 ]
[ m + 4 m - 6 ]
[--------------]
[863138 (m - 2)]
[ ]
[ 2 ]
[m - 4 m - 14 ]
[--------------]
[332862 (m - 2)]
[ ]
[ 12 ]
[--------------]
[265151 (m - 2)]
And the linear system and solution sol1 can be put in terms of three variables,
XYZ := Vector([ x, y, z ]):
M . XYZ = V;
[ 863138 x - 332862 y - 530302 z ] [ 8 ]
[ ] [ ]
[-530302 m z + 1726276 x - 665724 y] = [ -8 ]
[ ] [ ]
[ -863138 x - 998586 y - 1060604 z ] [-4 m]
Equate(XYZ, sol1);
[ 2 2 ]
[ m + 4 m - 6 m - 4 m - 14 12 ]
[x = --------------, y = --------------, z = --------------]
[ 863138 (m - 2) 332862 (m - 2) 265151 (m - 2)]
Now, another way, using three explicit equations,
reduc := LUDecomposition(Matrix([M,V]),output=U);
[863138 -332862 -530302 8 ]
[ ]
[ 0 -1331448 -1590906 -4 m + 8]
[ ]
[ 0 0 -530302 m + 1060604 -24 ]
GenerateEquations(reduc, convert(XYZ,list));
[863138 x - 332862 y - 530302 z = 8,-1331448 y - 1590906 z = -4 m + 8,
-530302 (m - 2) z = -24]
sol2 := solve(%,[x,y,z]);
[[ 2 2 ]]
[[ m + 4 m - 6 m - 4 m - 14 12 ]]
[[x = --------------, y = --------------, z = --------------]]
[[ 863138 (m - 2) 332862 (m - 2) 265151 (m - 2)]]
Notice that sol1 and sol2 agree. Note also that substituting m=2 into the result reduc above would produce an inconsistent system where the 3rd row down represented the impossible 0*x + 0*y + 0*z = -24.