I am asked to give 3 plane equation where the third plane will passes through the intersection of the first 2 planes and parallel to y axis.
I came up with 2 plane equation which is also parallel to y axis. x+0y+z-1 = 0 and x+0y+2z-2 = 0
subtract the first equation by the second i got z-1= 0 so z = 1 plugging in to the first equation it becomes x+1-1 = 0 so x = 0
Therefore the point will be (0,0,1)
I then plug it into the formula A1x + B1y + C1z + D1 + k(A2x + B2y + C2z + D2) = 0
It becomes 1(0)+0(0)+1-1+k(1(0)+0(0)+2(1)-2) = 0
but that makes the K equal to zero and so i cannot come up with the equation of the third plane.
The first two planes you have chosen have the line $x=0, z=1$ as their intersection, as you have shown (where $y$ is arbitrary).
Therefore you need an equation of the form $Ax+Cz=D$ to get a plane which is parallel to the y-axis, and you want to choose $A, C, D$ so that this plane contains all points of the form $(0,y,1)$ for any $y$.