Now that I THINK I better understand your question, thanks to @David White and @JMac
For the three cases:
The weight will be a minimum under water
The weight will be a maximum out of the water in a vacuum.
The weight out of the water under normal atmospheric temperature and pressure will be very, very slightly less than the weight in a vacuum.
For case 2, the only force acting on the ball is the downward force of gravity, $mg$. That’s the maximum force the scale will experience.
For case 3, in addition to the downward force of gravity the ball will experience an extremely small upward buoyancy force due to atmospheric air. That upward force will equal the weight of the air displaced by the ball, which would be the density of the air at standard atmospheric temperature and pressure times the volume of the ball. This would be a very, very small force. It would take an extremely sensitive scale to read this difference between the downward force of gravity and the upward buoyant force of air. If the upward buoyant force were greater than the downward force of gravity, such as for a helium inflated balloon, there would be no reading at all as the ball will rise.
For case 1, the upward buoyant force equals the weight of the water displaced by the ball. This upward force is much, much greater than case 3. The scale will read a minimum.
Hope this helps.