Ok, here is what I am thinking as an approach. Below I show the transformer connections for a Dyn bank that is standard connected per IEEE (low-side lags by 30° and assumes A-B-C rotation & subtractive polarity transformers).
With your low-side c-phase open circuited, no current can flow in the primary C-phase winding (neglecting magnetizing current). That constraint also prevents current flow in the primary A-phase winding as well in this particular case. So, the only current that can flow on the primary side is in the B-phase winding (the one in middle). As such, the only secondary winding that can have current flow is the b-phase winding.
So, I think you can simply reflect that b-phase load (\$5\Omega\$) to the primary and then forget about the transformer. Your problem now being reduced to a primary A-B fault with resistance (adding in your winding impedance data as desired).
Below is an example of sequence network connections for a B-C fault with resistance. You would shift angles for the A-B fault.
Note: Both images are from my lecture notes on symmetrical components.
Additional comments: If the secondary current of one of the 3 two-winding transformations is zero, \$I_S=0\$, then the primary current for that particular two-winding transformation is zero as well, \$I_P=0\$ (neglecting magnetizing current). Another way to look at it, if the secondary is open-circuited then it's load impedance is \$\infty\$...which when reflected to the primary winding by turn ratio squared is still \$\infty\$, an open circuit.
Also, the example phase-phase fault calculation in symmetrical components is done in per-unit (e.g. \$V_{BASE}=13.8\text{kV}, S_{BASE}=100\text{MVA}, \text{and} Z_{BASE} = \frac{13.8^2}{100}=1.904\Omega\$ so the \$5\Omega\$ resistance converted to per-unit is \$2.625\ \text{pu } \Omega\$. You can work it easier in your case all in actual units (Volts, Amps, Ohms) without the bother of converting to per-unit.
