#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}#
