Consider three data items D1, D2 and D3 and the following execution schedule…

To determine whether the schedule is serializable, we construct the precedence graph.

A schedule is serializable only if the precedence graph is acyclic.

Conflicting operations:

• T1 and T2 on D1:
  T1 writes D1 before T2 writes D1
  ⇒ T1 → T2

• T2 and T3 on D2:
  T2 writes D2 before T3 writes D2
  ⇒ T2 → T3

• T3 and T1 on D2:
  T3 writes D2 before T1 reads/writes D2
  ⇒ T3 → T1

Now the precedence graph contains:

T1 → T2
T2 → T3
T3 → T1

This forms a cycle:

T1 → T2 → T3 → T1

Since the precedence graph contains a cycle, the schedule is not conflict serializable.

Hence, the correct answer is Option D.

Also different cycle can be present.