Try:
eq = Rationalize[{m1, m2, m3}, 0];
Solve[eq, {x, y, z}]
(*{{x -> -(2986243790645883105455758358098195118556533001223336622172815\
95320218030787297467003699784937656128630985195404293779345/
122509485374294293794510320145189527584750827993252931995775611287\
31463104731917954143845606655037),
y -> 360193320714837502461108755235814898426025986503897780740376984\
7336638393825822545137993153228989027658267755878411291/
1225094853742942937945103201451895275847508279932529319957756112873\
1463104731917954143845606655037,
z -> 243238456696344063017037725294492493070825115410091835761432292\
011917023278886765613559961661062168696584881661936232099735/
1225094853742942937945103201451895275847508279932529319957756112873\
1463104731917954143845606655037}}*)
% // N(*Numeric solution:*)
(*{{x -> -2.43756*10^22, y -> 2.94013*10^20, z -> 1.98547*10^25}}*)
Solve is symbolic solver need exact values.
Is there any direct command to solve these equations? If someone does not use the rationalize command.
Also if I want to use the ans and want to find the value of “3x+2y”. It will simply show 3x+4y in output not calculate the numeric value of it
If your problem is linear then this
{vector,matrix}=CoefficientArrays[{m1,m2,m3},{x,y,z}];
solution=LinearSolve[matrix,vector]
will extract your coefficients from your equations and solve the linear system without needing to know or use Rationalize or N, and this then displays x,y,z
{2.43756*^22,-2.94013*^20,-1.98547*^25}
and then this will substitute that solution into your 3x+2y expression
ReplaceAll[3x+2y,Thread[Rule[{x,y,z},solution]]]
which displays the result
7.25388*^22
If you manually extract the coefficients from your equations and or insert the solution into your expression then you could do this without needing to know most of these functions.
Your equations are very poorly conditioned. The output can be meaningless:
{vector, matrix} = CoefficientArrays[{m1, m2, m3}, {x, y, z}];
matrix // Inverse
Solve in general handles numeric values as well.