Final answer:
The equation of the plane through the given points and perpendicular to the given plane is 5x + 4y - 2z + 18 = 0, which does not match any of the provided answer choices.
Step-by-step explanation:
The student is asking about finding the equation of a plane that passes through two points and is perpendicular to another plane. The given points are (0,-2,5) and (-1,3,1), and the equation of the other plane is 2z=5x+4y, which can be rewritten in normal form as 5x + 4y - 2z = 0. This indicates that the normal vector to the given plane is (5, 4, -2). For a plane to be perpendicular to this one, it must share the same normal vector. Therefore, we use the normal vector and one of the given points to determine the desired plane equation using the point-normal form of a plane equation, which is A(x - x1) + B(y - y1) + C(z - z1) = 0, where (A, B, C) is the normal vector and (x1, y1, z1) is a point on the plane.
By substituting the normal vector and one of the points, we can calculate the equation of the plane as:
5(x - 0) + 4(y + 2) - 2(z - 5) = 0
which simplifies to:
5x + 4y - 2z + 8 + 10 = 0
resulting in the equation of the plane:
5x + 4y - 2z + 18 = 0.
None of the answer choices provided in the question matches this equation, suggesting a potential typo or error in the question as presented.