Let $$p$$, $$q$$, $$r$$ be three statements such that the truth value of $$(p \wedge q) \to (\sim q \vee r)$$ is $$F$$. Then the truth values of $$p$$, $$q$$, $$r$$ are respectively:

We have the compound statement $$ (p \wedge q) \to (\sim q \vee r) $$ whose truth value is given to be $$F$$ (false).

First, recall the truth rule for an implication. The statement $$A \to B$$ is false only in one situation: when the antecedent $$A$$ is true and simultaneously the consequent $$B$$ is false. In every other case the implication is true.

Here the antecedent is $$p \wedge q$$ and the consequent is $$\sim q \vee r$$. Because the whole implication is false, we must have

$$ p \wedge q = T \quad \text{and} \quad \sim q \vee r = F. $$

Now we analyse each part in turn.

From $$ p \wedge q = T $$, the truth table of conjunction tells us that a conjunction is true only when both components are true. Hence

$$ p = T \quad \text{and} \quad q = T. $$

Next, look at the consequent $$\sim q \vee r$$. We already know $$q = T$$, so its negation is

$$ \sim q = \sim T = F. $$

The consequent simplifies to

$$ \sim q \vee r = F \vee r. $$

The truth table for a disjunction states that $$A \vee B$$ is false only when both $$A$$ and $$B$$ are false. We already have $$F$$ in place of $$A$$, so to make the entire disjunction false we must also have

$$ r = F. $$

Collecting the results, we find

$$ p = T, \; q = T, \; r = F. $$

Checking the option list, this matches Option A.

Hence, the correct answer is Option A.