The voltage drop on r_3 is increasing with an increase of r_2.
With increased resistance r_2, the combined resistance of parallel combination of r_2, r_3 and r_4 increases. This, in turn decreases the current going through the circuit according to the following calculations.
Let R be a combined resistance of the parallel combination of r_2, r_3 and r_4.
Then, according to the laws for parallel circuits,
1/R = 1/r_2 + 1/r_3 + 1/r_4
Obviously, R is increasing with an increase of r_2.
The combined resistance of an entire circuit is R+r_1, and it's also increasing with an increase of r_2.
Therefore, the current in the circuit is decreasing since it's equal to
I = V/(R+r_1),
where V is the voltage of the source of electricity and I is the current.
The voltage drop on the resistor r_1, which is equal to V_1=I*r_1, becomes smaller with an increase of r_2 since the current is decreasing. Since combined drop of the voltage must be equal to V, the voltage drop on a parallel combination of r_2, r_3 and r_4 is increasing. This voltage drop is the same for all three resistors r_2, r_3 and r_4 since they are connected in parallel. So, the voltage drop on r_3 is increasing with an increase of r_2.
Quantitatively, the voltage drop on each and all parallel resistors (including r_3, of course) equals to
V_p = I*R = (V*R)/(R+r_1) = V/(1+r_1*1/R) = V/(1+r_1*(1/r_2 + 1/r_3 + 1/r_4)
From the above formula we see that increase of r_2 causes decrease of denominator and, hence an increase of V_p.
