Final answer:
To find the equation of the plane that passes through a point and is perpendicular to two given planes, we can use the cross product of the normal vectors.
Step-by-step explanation:
To find the equation of the plane that passes through the point (1, 5, 1) and is perpendicular to the planes 3x + y - 3z = 3 and x + 5z = 7, we can use the normal vector of the given planes.
The normal vectors of two planes are perpendicular to the plane itself. Therefore, the cross product of the normal vectors would give us the direction vector of the plane we are looking for.
By finding the cross product of the normal vectors, we get the equation of the plane as: 16x + 3y - 3z = 38.