Consider the following grammar $G$, with $S$ as the start symbol. The grammar $G$ has three incomplete productions denoted by (1), (2), and (3). $$S \rightarrow daT \mid \underline{\quad (1) \quad}$$ $$T \rightarrow aS \mid bT \mid \underline{\quad (2) \quad}$$ $$R \rightarrow \underline{\quad (3) \quad} \mid \epsilon$$ The set of terminals is $\{a, b, c, d, f\}$. The FIRST and FOLLOW sets of the different non-terminals are as follows. $$\text{FIRST}(S) = \{c, d, f\}, \text{ FIRST}(T) = \{a, b, \epsilon\}, \text{ FIRST}(R) = \{c, \epsilon\}$$ $$\text{FOLLOW}(S) = \text{FOLLOW}(T) = \{c, f, \$\}, \text{ FOLLOW}(R) = \{f\}$$ Which one of the following options CORRECTLY fills in the incomplete productions?

To determine the correct productions, we analyze the given FIRST and FOLLOW sets:

1. Finding Production (2):

   • We are given $\text{FIRST}(T) = \{a, b, \epsilon\}$.

   • The productions for $T$ are $T \rightarrow aS \mid bT \mid (2)$.

   • For $\epsilon \in \text{FIRST}(T)$, the production (2) must be $T \rightarrow \epsilon$.

   • This eliminates options (C) and (D).

2. Finding Production (1):

   • We are given $\text{FIRST}(S) = \{c, d, f\}$.

   • The productions for $S$ are $S \rightarrow daT \mid (1)$.

   • Here, $\text{FIRST}(daT) = \{d\}$. Therefore, $\text{FIRST}((1))$ must contain $\{c, f\}$.

   • If $(1)$ is $S \rightarrow Rf$:

     $$\text{FIRST}(Rf) = (\text{FIRST}(R) \setminus \{\epsilon\}) \cup \{f\} = \{c\} \cup \{f\} = \{c, f\}$$

     This gives $\text{FIRST}(S) = \{d\} \cup \{c, f\} = \{c, d, f\}$, which matches the given set.

   • If $(1)$ were $S \rightarrow fR$, then $\text{FIRST}(fR) = \{f\}$, yielding $\text{FIRST}(S) = \{d, f\}$, which is missing $c$.

   • Thus, $(1)$ must be $S \rightarrow Rf$.

3. Finding Production (3):

   • With $(3)$ being $R \rightarrow cTR$, $\text{FIRST}(R) = \{c, \epsilon\}$ is satisfied.

Therefore, the correct choice is (A).