This is where the importance of the loads having different power factors come into play. You cannot just add up the currents, but also must look at the phase relationships between the currents.
I you have a 240V delta system, you have 240V between each of the phases separated by 120 degrees. So you can assign system voltages of Vab=240<0, Vbc=240<-120 and Vca=240<-240.
You have (1) 240V, 3ph load and (1) 240V single phase load. In my mind, I simplify the three phase load into (3) single phase loads balanced between the phases. So, you have 1 load between A-B, two loads between B-C and one load between C-A (that's my configuration, yours has two loads between A-B, but that doesn't change anything.
We know that the 3 phase load has a power factor of 0.9. From this we know that the load current lags (or leads) the load voltage by an angle of 25.84 deg. So for the load current of the load between A-B, we have Iab=S/Vab = 18.333kVA/240V = 76.39A. But the load current is 25.84 degrees out of phase from the voltage, so we would note this as Iab=76.39<-25.84. Likewise, the load currents for the other two parts of the 3 phase load would be Ibc=76.39<-145.84 and Ica=73.39<-265.84.
For the single phase load (connected between B-C in my calc.) we have a 70kVA, 240V load with a unity power factor. The current for this load Ibc2=70Kva/240V=291.67A. Since it is unity, the angle of the load current Ibc2 is the same as the angle of Vbc, so Ibc2=291.67<-120.
To solve the line currents Ia, Ib and Ic, you would use KCL (with vector math):
Ia = Iab - Ica
Ib = Ibc + Ibc2 - Iab
Ic = Ica - Ibc - Ibc2.
I think this will give you the values I have indicated.