If the equation of a plane is given by $ax+by+cz+d=0$, then the normal vector for that plane is $(a,b,c)$, if $a$, $b$, $c$ are not all $0$. Since every line contained in a plane is perpendicular to the normal vector, you can cross the two normal vectors of $x-y=1$ and $y+2z=3$ to find the vector in the direction of the line of intersection, which is contained in your desired plane. The reason for crossing the normals is that the line of intersection is contained in both planes, and thus perpendicular to both normal vectors.
Now two planes are perpendicular if their normal vectors are perpendicular, so to find the normal vector of your desired plane, you can cross the vector in the direction of the line of intersection, and the normal vector of $x+y-2z=1$.
Lastly, a plane is determined by a point in the plane and a normal vector, so all that remains is to find a point in the plane. You should be able to find one in the line of intersection.
Edit: Your situation looks something like this at first. Maybe it'll help in visualizing the space in which you're working.