Solution:
\( \displaylines{\text{Equating z we get}\\ x+y-1=1-x-2y\\ \text{which yields}\\ 2x+3y=2\\ y=\frac13\left(2-2x\right)\\ \text{To express as a vector, we write}\\ \lt x,y,z\gt=\lt x,\frac13\left(2-2x\right),x+y-1\gt\\ \text{Now substitute }x=t\text{ and make appropriate substitutions for }y\text{ and }z\\ \lt t,\frac23-\frac23t,t+\frac23-\frac23t-1\gt=\\ \lt0,\frac23,-\frac13\gt+t\lt1,-\frac23,\frac13\gt} \)