Phototransistors in parallel
When photo-transistors are connected in parallel, the way that is shown in your schematic, the \$V_{ce}\$ of each transistor will be equal, (assuming negligible voltage drop in the wires between the photo-transistors).
As a result, the photo-transistor that is most conductive will dominate the circuit. If you are familiar with how open-collector output pins may be &quot;wire-or&quot;ed together, you will understand this aspect of your circuit.
If this is not behavior that you want, then your circuit must be altered.
If you want to measure the average light intensity, perhaps the best way would be to using a summing amplifier, with each photo-transistor contributing to an input.
Edit:
Kevin White's answer states:

The problem is that the first transistor to be illuminated brings the output voltage to close to the supply, it is not possible for the output voltage to go higher than that so the second and subsequent transistors cannot raise the voltage any higher.


If you reduce the value of the 500k resistor so that a single transistor produces a small voltage much less than the supply (say 500mV) illumination on additional transistors will increase the voltage by the same amount.

I think this is clearer and more accurate than my own answer. Please change the accepted answer to Kevin White's (that is, if you agree).