Sending Voltage is higher than receiving voltage: Parallel Load Problem
Load one is \$S= 20\$ kW and \$+j~26\$ kvar.

Load two is \$S=60\$ kW and \$+j~60\sqrt{({1\over 0.8})^2 - 1}\$ kvar.

Load three is \$S=90\$ kW and \$-j~100\sqrt{1-0.9^2}\$ kvar.

Now you can calculate the current using \$({S\over V})^* = I\$. You need to use the complex conjugate here!  (If you want to know why, long story please ask in comment. :-)  

The voltage drop (rise) is \$V_l = Z_l\cdot I\$, giving:

$$V_s = 4146 \angle 0^o+V_l.$$