Two schedules are conflict equivalent if we can derive one schedule by swapping the non-conflicting operations of the other schedule.
S1$$\begin{array}{|l|l|l|}\hline T1 & T2 & T3 \\\hline & R(A) & \\\hline & W(A) & \\\hline & & {\color{Blue} {R(C)}} \\\hline & {\color{Blue} {W(B)}} & \\\hline & & W(A) \\\hline & & W(C) \\\hline R(A) & & \\\hline R(B) & & \\\hline W(A) & & \\\hline W(B) \\\hline \end{array}$$Here, we can swap R(C) and W(B) since they are non-conflicting pair (since they are operating on different data items)
After swapping the schedule will become $T2 \rightarrow T3 \rightarrow T1$ $$\begin{array}{|l|l|l|}\hline T1 & T2 & T3 \\\hline & R(A) & \\\hline & W(A) & \\\hline & W(B) & \\\hline & & R(C) \\\hline & & W(A) \\\hline & & W(C) \\\hline R(A) & & \\\hline R(B) & & \\\hline W(A) & & \\\hline W(B) \\\hline \end{array}$$
S2$$\begin{array}{|l|l|l|}\hline T1 & T2 & T3 \\\hline & & {\color{Blue} {R(C)}} \\\hline & R(A) & \\\hline & W(A) & \\\hline & W(B) & \\\hline & & W(A) \\\hline R(A) & & \\\hline R(B) & & \\\hline W(A) & & \\\hline W(B) & & \\\hline & & {\color{Blue} {W(C)}} \\\hline \end{array}$$ Here, we can swap and write R(C) after performing T2 operations:- R(A), W(A) and W(B) since each of them form non-conflicting pair with R(C) (since they are operating on different data items)
Also, we can swap W(C) and can execute it before all the T1 operations as each of the t1 operations are forming non-conflicting pair with W(C) (since they are operating on different data items)
After swapping the schedule will become $T2 \rightarrow T3 \rightarrow T1$ $$\begin{array}{|l|l|l|}\hline T1 & T2 & T3 \\\hline & R(A) & \\\hline & W(A) & \\\hline & W(B) & \\\hline & & R(C) \\\hline & & W(A) \\\hline & & W(C) \\\hline R(A) & & \\\hline R(B) & & \\\hline W(A) & & \\\hline W(B) \\\hline \end{array}$$
S3$$\begin{array}{|l|l|l|}\hline T1 & T2 & T3 \\\hline & R(A) & \\\hline & &R(C) \\\hline & & {\color{Blue} {W(A)}} \\\hline & {\color{Blue} {W(A)}} & \\\hline & W(B) & \\\hline & &W(C) \\\hline R(A) & & \\\hline R(B) & & \\\hline W(A) & & \\\hline W(B) & & \\\hline \end{array}$$Here, we can't swap the operations and make it as $T2 \rightarrow T3 \rightarrow T1$ because of the conflicting pairs W(A)and W(A)
$\therefore$ Option $D.$ S1 is conflict equivalent to S2, but not to S3 is the correct answer.