Equation of plane passing 2 intersecting planes and perpindicular to another

Find the equation of a plane that passes through the interesection of planes

x - z = 1

y + 2z = 3

and is perpendicular to x + y - 2z = 1

I already looked and understood the solution provided, however, I do not understand why my approach did not work. Here is what I did:

First, I solved for a point at the intersection of planes x - z = 1 and y + 2z = 3. One of the points happens to be (0, 5, -1)

Next, I take the plane x + y - 2z = 1 and find a vector that is coplanar to that particular plane. I do this by finding two points on the plane (0, 0, -1/2) and (1, 0, 0). With these two points, I form the vector <1, 0, 1/2>.

Finally, I use the vector equation of a plane to find the plane initially requested:

<1, 0, 1/2> . <x, y-5, z+1>

and get the cartesian equation: x + z/2 = -1/2

The answer in the solutions however is entirely different. (x + y + z = 4)

So what was the bad assumption that I made? Where did I go wrong? My approach seemed to make complete logical sense, unless of course I overlooked something. Thanks and any help is much appreciated.