x+y+z=1
x+y=2
let x=t
t+y+z=1
y=1-t-z
solving for x we get
x=3+t
now solving for y we get
y=-1-t
now solving forz we get
z=-1
therefore r(t)=3i-j-k+t(i-j)
Hmm. There's something wrong with your method if x = t and x = 3 + t. That's impossible, isn't it? I would suggest row reducing your augmented matrix thus:
You should get a one-parameter family of solutions. What do you get?