DFA for even number of As, odd number of Bs and precisely one C
Divide and conquer. You could design a DFA over the alphabet $\Sigma = \{A,B,C\}$  


for an even number of A, call it $M_1$
for an odd number of B, call it $M_2$
accepting exactly one C, call it $M_3$


Then design a new DFA equivalent to $M_1 \cap M_2 \cap M_3$ (by closure properties of regular languages).