A variant of Critical SAT in DP
[I made it into a proper answer b/c somebody gave it -1]

If any clause is allowed to be added, then the language is empty -- clearly to any satisfiable formula $F$ you can add a 3-clause $c$ made up of variables that do not appear in $F$: $F \cup \{ c \}$ will be satisfiable.

If the added clauses must use variables of $F$, then the language is in P.

Justification is as follows:

Take any $F \in L$, i.e. $F \in SAT$ and for any 3-clause $c$ on variables of $F$, $F \cup \{c\} \in UNSAT$. Say $c = l_1 \lor l_2 \lor l_3 \notin F$, where $l_i$ is a literal. Since $F \cup \{ c \}$ is UNSAT, all models of $F$ must have $l_i=0$ (for $i=1,2,3$) - because if some model had e.g. $l_1=1$, then it would satisfy $c$ and so $F \cup \{c\}$. Now, assume that there exists another clause $c&#39;$ that is exactly like $c$, but with one or more literal flipped and such that $c&#39; \notin F$, say $c&#39; = \neg l_1 \lor l_2 \lor l_3$. Then by the same argument all models of $F$ must have $l_1 = 1$. Thus, the necessary condition for $F \in L$ is that for each clause $c \in F$ there are exactly 6 other clauses in $F$ that use the three variables of $c$ -- lets call these 7-clause subsets of $F$ blocks. Note that each block implies a unique satisfying assignment to its variables. When this necessary condition is satisfied, $F$ is either uniquely satisfiable or unsatisfiable. The two cases can be distinguished by testing whether the assignments implied by the blocks of $F$ clash, which can clearly be done in linear time.