Understanding voltage drop on a transistor
If the transistor base current is adequate the transistor will be driven into saturation and there will be about 0.2 V between the collector and emitter. This is small enough that it can be ignored in many calculations such as this.
That leaves about 3 V across the resistor so I = 3/75 = 40 mA shared between the three LEDs. Generally direct paralleling of the LEDs is avoided because the currents will vary quite a lot depending on their individual forward voltages, Vf.
See what I've written in Variations in Vf and binning for more on the topic.