Determine an equation of a plane parallel to a coordinate plane that contains the points (3, 4, −5) and (−2, 8, −5).

To find the equation of a plane parallel to the coordinate plane and passing through two given points (3, 4, -5) and (-2, 8, -5), we can use the following method: Find the vector directed from the first point to the second. To do this, subtract the coordinates of the first point from the coordinates of the second point: Vector AB = (-2 - 3, 8 - 4, -5 - (-5)) = (-5, 4, 0). Since we are looking for a plane parallel to the coordinate plane, we want the normal vector of that plane to be parallel to the normal vector of the coordinate plane (0, 0, 1), since the z coordinate plane has a normal vector (0, 0, 1). Now we need to find the equation of the plane using the found normal vector and one of the points (let's use the point (3, 4, -5)): The plane equation is Ax + By + Cz + D = 0, where (A, B, C) is a normal vector and (x, y, z) is a point on the plane. Substitute the values: (-5)x + (4)y + (0)z + D = 0. Simplify the equation, given that C = 0: -5x + 4y + D = 0. Now we find the value of D by substituting the coordinates of the point (3, 4, -5): -5(3) + 4(4) + D = 0, -15 + 16 + D = 0, 1 + D = 0, D = -1. So, the equation of a plane parallel to the coordinate plane and passing through the points (3, 4, -5) and (-2, 8, -5) has the form: -5x + 4y - 1 = 0.