The correct option is A T, T, F
Given p→(∼q∨r) is false.
Option A) If p=T, q=T, r=F
so, T→(∼T ∨F)T→(F∨F)T→F
which is equal to F
Option B) If p=T, q=F, r=T
so, T→(∼F ∨T)T→(T∨F)T→T
which is equal to T
Option C) If p=F, q=T, r=T
so, F→(∼T ∨T)F→(F∨T)T→T
which is equal to T
Option D) If p=F, q=T, r=T
so, F→(∼T ∨T)F→(F∨T)F→T
is equal to T
Alternate solution:
x→y is F only when x is T and y is F.
∴ p→(∼q∨r) is false only when P is T and (∼q∨r) is F, which is possible only when both ∼q and r are F.
So, q is T and r is F.