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#.