To find the distance between a point and a plane, we can use the formula:

Distance = |ax + by + cz + d| / sqrt(a^2 + b^2 + c^2)

where the plane equation is ax + by + cz + d = 0 and (x, y, z) is the point.

Given planes:
1) 2x - 2y + z = 0
2) x - y + 2z = 4

We need to find the equation of the plane that passes through the point (1, 2, 2) and is perpendicular to these two planes.

Step 1: Finding the normal vectors of the given planes
The normal vector of a plane is given by the coefficients of x, y, and z in the plane equation. So, for the first plane, the normal vector is (2, -2, 1), and for the second plane, the normal vector is (1, -1, 2).

Step 2: Finding the cross product of the normal vectors
The cross product of two vectors gives a vector perpendicular to both of them. So, we can find the cross product of the normal vectors of the given planes to get the direction of the normal vector for our required plane.

Cross product: (2, -2, 1) x (1, -1, 2)
= [(2)(2) - (-2)(-1), (-2)(1) - (2)(1), (2)(-1) - (1)(-2)]
= (4 - 2, -2 - 2, -2 - 2)
= (2, -4, -4)

Step 3: Finding the equation of the required plane
We have the direction vector of the required plane, but we need a point that lies on the plane to determine its equation. We are given that the plane passes through (1, 2, 2), so we can use this point along with the direction vector to find the equation.

The equation of the plane is given by:
2(x - 1) - 4(y - 2) - 4(z - 2) = 0
2x - 2 - 4y + 8 - 4z + 8 = 0
2x - 4y - 4z + 14 = 0
x - 2y - 2z + 7 = 0

Step 4: Finding the distance between the point (52, 53, 57) and the plane
Using the formula mentioned earlier, we can find the distance between the point (52, 53, 57) and the plane x - 2y - 2z + 7 = 0.

Distance = |(2)(52) - 4(53) - 4(57) + 14| / sqrt(2^2 + (-4)^2 + (-4)^2)
= |104 - 212 - 228 + 14| / sqrt(4 + 16 + 16)
= |-322| / sqrt(36)
= 322 / 6
= 53.67

The square of the distance is (53.67)^2 = 2878.76, which is closest to 5202. Therefore, the correct answer is option A.