JacquesBBB, don't worry about Steverino's solution, he's trying to solve the much, much more difficult case where all the resistances are different (rather than all being equal to 1k)!

Steverino, the parallel resistance formula allows you take take a group of resistors which connects precisely the same two nodes, and replace them all with a single resistor (which connects the same two nodes as before.) For example, if we had rX going from node A to node B, and another resistor rY going from A to B as well, then we can delete those two resistors and replace it with a resistor of value (rX || rY) going between A to B.

The main problem with your solution is that you're applying this formula even though r6, r9 and r12 do not connect precisely the same two nodes. For example, r6 goes from B to A, but r9 goes from B to C. Unfortunately, you're simply not allowed to apply the parallel resistance formula in this case.

Now if (and only if) the resistances are all equal, and if you apply the reasoning showed by Brumby, then nodes A, C and b become connected (therefore they become the same node), and only then does the parallel resistance formula become applicable. But because we had to assume all the resistances are equal to pull this trick, you're not allowed to claim that your approach gives a valid formula when all the resistances are arbitrary/different. (And, as demonstrated in my earlier messages, your formula gives clearly wrong results in certain contrived arbitrary cases.)