Using Gaussian Elimination on the given system, we start from
#{(2xy+z=6),(x+2yz=1),(2xyz=0):}#
and use the system to set up the augmented matrix
#((2,1,1,,6),(1,2,1,,1),(2,1,1,,0))#
Our goal is a matrix of the form
#((1,0,0,,x),(0,1,0,,y),(0,0,1,,z))#
Note that from here, we may use only the following operations:

multiply a row by a constant

add a the product of a row and a constant to another row

switch the positions of two rows
To obtain a #1# in the first row, first column, we multiply the first row by #1/2# and obtain
#((1,1/2,1/2,,3),(1,2,1,,1),(2,1,1,,0))#
To obtain #0#s for the remainder of the first column, we add #1# times the first row to the second row, and #2# times the first row to the third row.
#((1,1/2,1/2,,3),(0,5/2,3/2,,2),(0,0,2,,6))#
We could solve for #z# and use backsubstitution at this point, but let's proceed with GaussJordan elimination. The next step is to get a #1# in the second row, second column. To do so, we multiply the second row by #2/5#.
#((1,1/2,1/2,,3),(0,1,3/5,,4/5),(0,0,2,,6))#
Again, we want #0#s for every other member of the column. To get this, we add #1/2# times the second row to the first row.
#((1,0,1/5,,13/5),(0,1,3/5,,4/5),(0,0,2,,6))#
As our next to last step, we want a #1# in the third column, third row. To obtain this, we multiply the third row by #1/2#.
#((1,0,1/5,,13/5),(0,1,3/5,,4/5),(0,0,1,,3))#
Finally, we eliminate the remaining nonzero elements of the third column by adding #1/5# times the third row to the first row, and #3/5# times the third row to the second row.
#((1,0,0,,2),(0,1,0,,1),(0,0,1,,3))#
And so we have
#{(x = 2),(y=1),(z=3):}#