If A $\cap$ B $\cap$ C = $\emptyset$, then the sum principle applies so |A $\cup$ B $\cup$ C| = |A|+|B|+|C|.
I think it would be true since there is nothing in common among A,B and C, but just wondering if there is any exceptions to this problem so it would be false?
This is false, because while $A \cap B \cap C$ might be empty, this does not imply that $A \cap B$ is empty (nor that $B \cap C$ and $A \cap C$ are).
Indeed the intersection of all three might be empty while each element of the union is counted twice when $|A| + |B| + |C|$ is summed.
Let A = {1,2}, B={2,3} and C={3,1} so |A| = |B| = |C| = 2. Then A $\cap$ B $\cap$ C = $\emptyset$, but $$A \cup B \cup C = \{1,2,3\} \implies |A \cup B \cup C|=3 \not= |A| + |B| + |C|$$
When you apply inclusion-exclusion in this case, you also need to consider the sizes of |A $\cap$ B|, |B $\cap$ C| and |C $\cap$ A|. I suspect the cause of the confusion is that you have generalized from the two set case, in a way that doesn't work in the three set case. The "sum principle" you refer to, is exactly the inclusion-exclusion principle applied to only two sets: $$|A \cup B| = |A| + |B| - |A\cap B| $$
If and only if A and B are mutually exclusive, so $A \cap B = \emptyset $, we have: $$|A \cup B| = |A| + |B| $$
But when three sets are involved, the inclusion-exclusion principle gives:
$$|A \cup B \cup C| = |A| + |B| + |C| - |A\cap B| - |B\cap C| - |C\cap A| + |A\cap B\cap C| $$
You can check why this works visually. Can you see why each subset of A $\cup$ B $\cup$ C is only counted once? For instance the part of A that does not intersect with B or C (i.e. $A \cap B' \cap C'$) is only counted by the first term. More complicated is $A \cap B \cap C$ (in the middle), which is included by the first three terms so it starts off as triple-counted. The next three terms take it away again, and the final term adds it back on, so by the end it has been counted once only.
Knowing that $A\cap B\cap C = \emptyset $ would only reduce this to:
$$|A \cup B \cup C| = |A| + |B| + |C| - |A\cap B| - |B\cap C| - |C\cap A|$$
In my example $A\cap B=\{2\}$, $B\cap C=\{3\}$ and $C\cap A=\{1\}$ which all have cardinality one. The formula correctly gives $3 = 2 + 2 + 2 - 1 -1 -1$.
Note that $A\cap B\cap C$ is all the elements that are in all three sets. It does not mean that $A\cap B=\emptyset$, or $B\cap C=\emptyset$.
Use the inclusion-exclusion principle
$$|A\cup B \cup C| = |A| + |B| + |C| - |A\cap B| - |A\cap C| - |B\cap C|
+ |A\cap B \cap C|.$$
