We are given the following information.
Resistor: R₁ = 400 Ω
Resistor: R₂ = 600 Ω
Resistor: R₃ = 500 Ω
We are asked to find the equivalent resistance of the given circuit.
Notice that the resistors R₁ and R₂ are in parallel and this parallel combination is in series with resistor R₃.
First, let us find the parallel resistance of R₁ and R₂.

Finally, let us find the series resistance of Rp and R₃.

Therefore, the equivalent resistance of the given circuit is 740 Ω.