 Solution

Take the single Boolean clauseThis is a 3-CNF formula. Suppose a 2-CNF formula $K$ were logically equivalent to $F$. Every Boolean clause $C$ of $K$ would then be a consequence of $F$.

Consider any nontautological Boolean clause $C$ containing at most two Boolean literals. Choose a Boolean valuation falsifying its Boolean literals. At least one of $P,Q,R$ does not occur in $C$; assign that Boolean variable true. The resulting Boolean valuation satisfies $F$ while falsifying $C$. This contradicts $F\models C$. An empty Boolean clause also cannot be a consequence of the satisfiable $F$.

Consequently every Boolean clause of $K$ would have to be a tautology. Their logical conjunction would be true under every Boolean valuation, whereas $F$ is false when $P=Q=R=0$. This proves that no such $K$ exists.

The argument even allows additional ordinary Boolean variables in $K$: at most two Boolean literals still leave one of $P,Q,R$ free to be made true. Introducing extra variables together with existential quantification would change the requested notion of logical equivalence, so it is not a way around the argument.