Think of the line to line voltages as a delta connection. When you go from a delta to a wye connection, your phase angle shifts by -30 degrees. The reason for this can easily be calculated when you show the addition of the A-N and B-N vectors to give you the A-B vector. I won't do it here, but do a search for "delta-wye phase shift" and you should find all the info you need. So that explains the phase angle shift, and the magnitude is of course the line voltage over the square root of 3.
Now, the key to the sequence is that you must remember that voltage/current vectors ALWAYS ROTATE COUNTER-CLOCKWISE. So use zero degrees as your reference. If the rotation is ABC, you'll see A come around first, then B, then C. If you freeze them when A is at zero, then B will be the next one to come around, and so it must be at -120 degrees (240 degrees). C will come around last, so it's sitting at 120 degrees.
So in the problem, AB is given to be at 0 degrees. For ABC rotation, this means BC is next and will be at -120, then CA at 120. If the rotation is changed to ACB, then you'll have AC at 0, CB at -120, and BA at 120.
Once you've got this figured out, you just perform your phase shifting as noted above and you can get your L-N voltages pretty easily. The current angles are simply their respective L-N voltage angles minus the impedance angle. In this case, the Cosine of the impedance angle will also give you your power factor.
Hope that helps, or at least didn't make things more confusing.
