Let P1 and P2 be the planes 2x-2y+z=0 and x-y+2z=4 respectively.
The normal vector to P1 is (2,-2,1) and to P2 is (1,-1,2).
The perpendicular plane P3 has to contain point (1,-2,1).
The equation of P3 can be written a(x-1)+b(y+2)+c(z-1)=0 being the scalar dot product of its normal vector (a,b,c) and the difference of the general point vector on P3 (x,y,z) and vector of point (1,-2,1).
The scalar product of P1 normal with P3 normal must be zero because they are perpendicular, since P1 and P3 are perpendicular. That is, (2,-2,1)•(a,b,c)=0. That is, 2a-2b+c=0.
Similarly P2 and P3 normals: (1,-1,2)•(a,b,c)=0, so a-b+2c=0. Doubling this: 2a-2b+4c=0, therefore, since 2a-2b=-c from the above equation c=0, and a=b, and a(x-1)+a(y+2)=0⇒x-1+y+2=0, x+y+1=0 is the equation of the plane P3, with the normal (1,1,0).
The magnitude of the normal (1,1,0) of P3 is √(1^2+1^2+0^2)=√2. If we take any other point (x,y,z) on the plane P3, we can work out the difference vector between this and the required point (1,2,2). We know that x+y=-1 for all points on P3, so n • r=-1. The dot product n • (r - r0) where r=(x,y,z) and r0=(1,2,2), gives the projection on to the normal:
n • (r - r0) = n • r - n • r0=-1-(1*1+1*2)=-4. We now have to divide this by the magnitude of the normal, √2, to give -4/√2=-4√2/2=-2√2. We need this as a magnitude, or absolute value,=2√2 or 2*2^0.5 or 2(2^0.5).