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.

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!