You seem to be confused on several points. In order to lie on "the intersection of the planes x−y+z=0 and x+y+2z=0" a point must satisfy both equations which is true for the origin but not, as tkhunny said, (3, 0, 0).
You also say "the" parameterization which is wrong- any curve has an infinite number of parameterizations. I presume you are asking for "a" parameterization.
One simple way is to note that if you add the two equations you eliminate "y" and get 2x+ 3z= 0. We could solve for x and write x= -(3/2)z, then take z itself as parameter so that x= -(3/2)t, y= x+ z= -(3/2)t+ t= -(1/2)t, z= t. Or take x itself as parameter: x= t, y= x+ z= t- (2/3)t= (1/3)t, z= -(2/3)t. Or, if you don't like fractions, observe that 2x+ 3z= 0 is satisfied by x= 3, z= -2 or any multiple of those: 2(3t)+ 3(-2t)= 6t- 6t= 0. So take x= 3t, y= x+ z= t, z= -2t.
(Taking t= 1 in the last parameterization, (3, 1, -2) lies on the intersection, not (3, 0, 0). As a check, 3- 1- 2= 0 and 3+ 1- 2(2)= 0 so (3, 1, -2) lies on both planes so on their intersection.)
