Thanks for your help! If you have time, maybe you could explain me the meaning of each input? From your answer and maybe from the one in the other block post:

m = {{1, 1, 1}, {1, 2, 3}, {1, 4, 9}};
b = {1, 2, 3};
abc = LinearSolve[m, b]
f = Dot[abc, {x, y, z}]
sol = Solve[f == 0, z]
Plot3D[z /. sol, {x, 0, 15}, {y, 0, 15}]

, because I would like to understand the system.

If you do not have time, it does not matter, because you have already helped me a lot!

Thanks for your answer! Is it possible to get this solution set with the parameters just using e.g. Solve[...] or so, why does it have to be so complicated?

Is there also a command for this, this matrix is represented as a plane equation? So: x=a+rb+tc (with r and t as parameters and x,a,b,c as vectors)

Solution for an underdetermined system is not unique:

A = {{6, 3, 2, 3, 4}, {4, 2, 1, 2, 3}, {4, 2, 3, 2, 1}, {2, 1, 7, 3, 
    2}};
B = {5, 4, 0, 1};
ns = NullSpace[A];
x0 = Sum[C[i]*ns[[i]], {i, 1, Length[ns]}];
x1 = LinearSolve[A, B];
x1 + x0
A . % == B // Simplify (*Solution are: TRUE *)
(*{-(3/2) + 3 C[1] - C[2], 2 C[2], -2 + 4 C[1], 6 - 14 C[1], 4 C[1]}*)
(* TRUE *)

or:

Table[C[i], {i, 1, Length[A[[1]]]}] /. 
 Solve[A . Table[C[i], {i, 1, Length[A[[1]]]}] == B, 
  Table[C[i], {i, 1, Length[B]}], MaxExtraConditions -> All] // Expand
A . %[[1]] == B // Simplify (*Solution are: TRUE *)
(*{{C[1], -3 - 2 C[1] + (3 C[5])/2, -2 + C[5], 6 - (7 C[5])/2, C[5]}}*)
 (* TRUE *)

Maple 2022.2 solution:

Regards M.I.