Why does a fault lead to increased reactive power demand?
The reactive power loss in any short, lossless transmission line is given by,

\$\Delta Q= \frac{P_{flow}^2+Q_{flow}^2}{V^2}\times X\$

where \$P_{flow}\$ and \$Q_{flow}\$ are the active and reactive power flow in the line.

Now let us analyse

Pre-fault

The power flowed from bus1 to bus2 is divided on 6 lines. Which means that the power flow in each line is 1/6 of total power (\$P_{total}\$), assuming that all lines having same parameters and length. This  \$\frac{1}{6} P_{total}\$ flowing in each line will not lead to much voltage drop and reactive power losses in each line.

Post-Fault

When three lines are tripped, the total power \$P_{total}\$ will be redistributed among all lines. Again, assuming that all lines having same parameters and length, the share of power flow in each line \$\frac{1}{3} P_{total}\$ which means double of power flow, and hence (from equation above) almost \$2^2=4\$ times more reactive power losses compared to pre-fault case.

Moreover, the voltage drop across the lines will increase as total impedance of the lines is significantly increased (doubled in case of same parameters and length) after fault occurred compared to pre-fault case.