Wye-delta equivalent resistance question
My previous answer was unhelpful to the OP, so I will attempt a different approach and delete the old answer.

Wye-delta transformations are mathematical and built on the same principle as Thevenin and Norton Equivalents - the superposition theorem for electric circuits.

I won't get into the details of the proof, but if you like it is explained fairly well here.  Fundamentally, it states that the response of a bilateral* linear** circuit with multiple sources can be determined by algebraically summing the responses of the circuit to each source, with the remaining sources replaced by characteristic impedances.

*response is independent of current direction
**containing no non-linear elements (capacitors, inductors, etc)

To do this transformation, we apply an ideal current to nodes 1, 2, and 3 (respectively) in each step of the transformation:

1) (I1-I2), -(I1-I2), 0
2) 0, (I2-I3), -(I2-I3)
3) -(I3-I1), 0, (I3-I1)

This is consistent with Kirchoff's Current Law, as I1 + I2 + I3 = 0.

By choosing this approach, we eliminate one of the nodes from each step, allowing us to compute the equivalent resistance across the remaining nodes as we would normal series and parallel resistances.  We then linearly sum those results to give a final equivalence.

In short, it is a choice backed by mathematical principles of superposition to make the problem inherently easier to solve.