The correct option is A T,F,F
We know that,p→(q∨r)⇒ ∼p∨(q∨r)
Now putting values
T,F,F⇒ ∼T ∨(F∨F)=F
F,F,F⇒ ∼F∨(F∨F)=T
F,F,T⇒ ∼F∨(F∨T)=T
T,T,F⇒ ∼T∨(T∨F)=T
Hence, T,F,F is correct.
Alternate Solution
we know that s→t is false iff s is true and t is false
So, p→(q∨r) will be false iff p is true and q∨r is false
∴ Truth values for (p,q,r)≡(T,F,F)