0x+3y+2z=4
2x−y−3z=3
2x+2y−z=7
((0,3,2quadquad|4),(2,-1,-3|3),(2,2,-1|7))~~((2,-1,-3|3),(2,2,-1|7),(0,3,2quadquad|4))~~
R_3hArrR_1
R_2=R_2-R_1
~~((2,-1,-3|3),(0,3,2quad|4),(0,3,2quad|4))~~((2,-1,-3|3),(0,3,2quad|4),(0,0,0quad|0))~~
R_3=R_3-R_2
R_2=1/3*R_2
~~((2,-1,-3|3),(0,1,2/3quad|4/3),(0,0,0quad|0))~~((2,0,-7/3|13/3),(0,1,2/3quad|4/3),(0,0,0quad|0))~~((1,0,-7/6|13/6),(0,1,2/3quad|4/3),(0,0,0quad|0))
R_1=R_1+R_2
R_1=1/2*R_1
x-7/6z=13/6
y+2/3z=4/3
Last row has all zeros which means we have to substitute for parameter. z seems to be the best way to go.
z=p (p as parameter)
x-7/6p=13/6quad=>quadx=13/6+7/6p=(13+7p)/6
y+2/3p=4/3quad=>quady=4/3-2/3p=(4-2p)/3
{((13+7p)/6, (4-2p)/3, p), p in RR}