x = 3a + 2b, y = b + c, z = -a + b + 2c
Ansatz:
mx + ky + nz = 0
m(3a + 2b) + k(b+c) + n(-a + b + 2c) = 0
(3m -n)a + (2m + k+n)b +(k +2n)c = 0 da a,b,c lin. unah. ==>
3m - n = 0 → m = 1/3 n
2m + k + n = 0
k + 2n = 0 → k = -2n
2/3 n - 2n + n=0
-1/3 n = 0 =====> n=0
==> m=0
==> k =0
Daher sind x,y und z lin unabh.