Use this Venn diagram:



We want to find y=?

a survey of 220 families showed that: 
s+t+u+v+w+x+y+z=220
63 had a dog; 
s+t+v+w=63
64 had a cat;
t+u+w+x=64 
28 had a dog and a cat; 
t+w=28
84 neither had a cat nor a dog and in addition did not have a parakeet; 
z=84
6 had a cat, a dog and a parakeet. 
w=6

how many families had a parakeet only?  That's y.

     s t u v w x y z
     | | | | | | | |
(1)  s+t+u+v+w+x+y+z=220
(2)  s+t  +v+w      = 63
(3)    t+u  +w+x    = 64
(4)    t    +w      = 28
(5)                z= 84
(6)          w      =  6

Substitute w=6 in eq. (4) to get t=22
Substitute t=22, w=6, z=84 in the first three 
equations and simplify:

     s t u v w x y z
     | | | | | | | |
(1)  s  +u+v  +x+y   =108
(2)  s    +v         = 35
(3)      u    +x     = 36

Subtract eqs. (2) and (3) from eq. (1) 

and get y = 37

Edwin

We have a universal set of 220 families and three its basic subsets

    d  of 63 families having a dog;

    c  of 64 families having a cat;

    p  of unknown number of families having a parakeet.


The number of families in the union (d U c) is  63 + 64 - 28 = 99.


The set of families that neither have a cat nor a dog  is the complement to the set (d U c),
and the number of such families is  220 - 99 = 121.


We are told, that the number of families that "in addition" do not have parakeet is 84.


From it, we conclude that  121-84 = 37  families do have a parakeet, but have neither a cat nor a dog.


So, the answer to the problem's question is 37 families.

    "6 had a cat, a dog and a parakeet. "