{a^n b^n c^n | n>=1} - PDA
Yes you missed something ,
Lets review how your PDA works :

we push  a for each a we read
we pop 2as for each b we read , then when stack is empty we push b for each b we read
we pop b for each 2cs we read

Your PDA will accept strings of form a2nbn+mc2m
for ex , a4b5c6 , here n = 2 , m = 3
You push 4 as , then for each 2as you pop b , you pop 2bs , and 3 bs remain which are pushed to stack !!!
Hopefully you can see the problem here , no one told you that you are half way bs when you finished popping as
You continue , for each of the 6 cs you pop a b then you accept
You expect the string in the form a2nbn+mc2m , you compare  a2nbn then  bmc2m , of course if n = m you are correct , but there is no way to ensure this