6 phase forked star transformer
I understand your problem and the confusion caused by the reference diagram provided.  I always try to align my diagrams with the phasors when I draw them at \$\pm 120^\circ\$, otherwise I draw them all horizontally to show the ampere-turn balance.

Let us assume the top-left delta is correct.  I will then draw the phasors $$\DeclareMathOperator{\kV}{~kV}\DeclareMathOperator{\V}{~V}U_{CB} = 11\kV\angle 0^\circ,$$

$$U_{BA} = 11\kV\angle 120^\circ$$ and $$U_{AC} =11\kV\angle -120^\circ =11\kV\angle 240^\circ.$$

This implies looking at the top-right diagram showing the interconnections and implied core locations by the angle drawn. Therefore from primary to secondary:  $${1\over100} U_{CB} = U_{nh}  = U_{eg} = U_{di} = 110\V\angle 0^\circ,$$
$${1\over100} U_{BA} =  U_{ng} = U_{ci} = U_{bh} = 110\V\angle 120^\circ$$
and $$U_{ni} = U_{fg}= U_{ah} = 110\V\angle 240^\circ$$

The secondary line voltage is then for example $$|U_{bh}-U_{ah}| = |110\V\angle 120^\circ - 110\V\angle 240^\circ| = {\sqrt3\cdot110\V}.$$

(Prepare to get totally lost if you try to use the phasor diagrams provided. :-)