If the truth value of the Boolean expression $$((p \vee q) \wedge (q \rightarrow r) \wedge (\sim r)) \rightarrow (p \wedge q)$$ is false, then the truth values of the statements $$p$$, $$q$$, $$r$$ respectively can be:

We begin with the compound implication $$\big((p \vee q) \wedge (q \rightarrow r) \wedge (\sim r)\big) \rightarrow (p \wedge q).$$

The standard truth table rule is: an implication $$A \rightarrow B$$ is false only when $$A$$ is true and $$B$$ is false. In every other case it is true.

Because the whole expression is stated to be false, we must therefore have

$$\big((p \vee q) \wedge (q \rightarrow r) \wedge (\sim r)\big)=\text{True}$$

and simultaneously

$$ (p \wedge q)=\text{False}. $$

Let us analyse each part in turn.

First, the consequent $$p \wedge q$$ is false. A conjunction is false whenever at least one of its components is false, so at least one of $$p$$ or $$q$$ must be false.

Next, the antecedent is a conjunction of three statements, and we are told the conjunction is true. Therefore each of its three factors must be true:

1. $$(p \vee q)=\text{True}.$$

2. $$(q \rightarrow r)=\text{True}.$$

3. $$(\sim r)=\text{True}.$$

From the third condition $$\sim r=\text{True}$$ we obtain immediately

$$ r=\text{False}. $$

Now consider the second condition $$q \rightarrow r=\text{True}.$$ Recall the logical equivalence $$q \rightarrow r \equiv (\sim q) \vee r.$$

Substituting the already known value $$r=\text{False},$$ we get

$$ (\sim q) \vee \text{False}=\text{True}. $$

The only way this disjunction can be true is if $$\sim q=\text{True},$$ hence

$$ q=\text{False}. $$

Finally, look at the first factor $$p \vee q=\text{True}.$$ We have just found $$q=\text{False},$$ so for the disjunction to be true we must have

$$ p=\text{True}. $$

Collecting our results, we have

$$ p=\text{True}, \quad q=\text{False}, \quad r=\text{False}. $$

Written in the order $$p,\,q,\,r$$ this is $$T F F,$$ which matches Option B.

Hence, the correct answer is Option 2.