Homework? (Doesn't seem like the kind of problem one does for recreation.)
What do you mean "the plane"? There are an infinite number of planes that satisfy the condition- take anyone plane that does and move it "parallel to itself" along the line of intersection of the first two planes. In fact, since we are requiring that the plane is perpendicular to a plane rather than a line- there are an infinite number of planes perpendicular to a given plane.
The first thing you should do is find the line of intersection of the two planes: x-z = 1 and y +2z =3. We can write a parametric equation for the line by taking z= t, the parameter, and solving the other two equations for x and y in terms of z: x= z+1 and y= 3- 2z so the parametric equations are x= t+1, y= 3- 2t, z= t.
Given that the plane is perpendicular to x+ y- 2z= 1, it's normal vector is perpendicular to the normal vetor of that plane, <1, 1, -2>.
In other words, the equation of the plane must be of the form
a(x-x0)+ b(y-y0)+ c(z-z0)= 0 where a+b-2c= 0 so that if b and c are any two numbers then a= 2c- b gives a plane perpendicular to x+ y+ 2z= 1.
To guarantee that the plane passes through the given line, we can make (x0,y0,z0) a point on the line. Take t= 0 so that (1, 3, 0) is a point on the line. Take c= 1, b= 1 so that a= 2(1)-1= 1. Then the plane 1(x-1)+ 1(y-3)+ 1(z-0)= x-1+ y- 3+ z= 0 or x+ y+ z= 4.
x+ y+ z= 4 crosses the line of intersection of the two planes (at (1, 3, 0)) and is perpendicular to the plane x+ y+ 2z= 1 (it's normal vector is <1, 1, 2> and <1, 1, 2>*<1, 1, -2>= 1+ 1- 4= 0).
But my point is that that is one of many such planes. Taking b, c, and t to be any real numbers, the plane
(2c-b)(x-(t+1))+ b(y- (3-2t))+ c(z- t)= 0 passes through the line of intersection and is perpendicular to the given plane.