Finding Different Paths and Critical Path :

A$-$B$-$D$-$F $=3+3+3+$

A$-$C$-$E$-$F $=3+4+4+\_=12($i$.$e$.$ Duration for Activity F is $1(1*1))$

Therefore

x $=1$

Paths:

A$-$B$-$D$-$F $=3+3+3+1=10$

A$-$C$-$E$-$F $=3+4+4+1=12($Critical Path$)$

Explanation:

Based on the given Network paths and Critical Path $.$ The path which is having Maximum Duration is Critical Path.