Each Schedule works on different data item so no need to consider that. here ONLY A is shared between all three schedule and only operation is Blind write in all three schedule hence no need to check first read and write – read dependency only rule to keep in mind is last write on A.
Let assume T-1 has last write on A so fix T-1 at last position to follow last write. now T-2 and T-3 can be arranged in 2 ways and order doesn’t matter here because no case of dirty read on A .
D is correct