Let's assume we have a chess board of $4\times4$ and $4$ different rooks. We are not allowed to place the rooks on the white main diagonal. So we have $12$ spots to place the rooks. We want to know how many ways exist that in each column is at least one rook.
Let be $\Omega:=\{(\omega_1,\omega_2,\omega_3,\omega_4)\mid 1\leq\omega_i\leq 12\text{ and }\omega_i\neq\omega_j\text{ if } i\neq j \}$ where each $\omega_i$ denotes the position of rook $i$. The cardinality of the sample space is $|\Omega|=12\cdot11\cdot10\cdot9$.
Let be $A_i$ the set where in the $i$-th column is at least one rook. So considering baisc counting principles and that in a column three spots are allowed we get $|A_i|=4\cdot3\cdot11\cdot10\cdot 9$.
Let be $A_i\cap A_i$ where $1\leq i\neq j\leq 4$ the set where in the $i$-th and $j$-th column is at least one rook. So $|A_i\cap A_j|=4\cdot3\cdot3\cdot3\cdot10\cdot 9$.
Let be $A_i\cap A_i\cap A_k$ where $1\leq i\neq j\neq k\leq 4$ the set where in the $i$-th, $j$-th and $k$-th column is at least one rook. So $|A_i\cap A_j\cap A_k |=4\cdot3\cdot3\cdot3\cdot2\cdot 3\cdot 9$.
Let be $A_i\cap A_i\cap A_k\cap A_m$ where $1\leq i\neq j\neq k\neq m\leq 4$ the set where in the $i$-th, $j$-th, $k$-th and $m$-th column is at least one rook. So $|A_i\cap A_j\cap A_k \cap A_m|=4\cdot3\cdot3\cdot3\cdot2\cdot 3\cdot 3$.
Applying the inclusion/exclusion principle yields: $$ |A_i\cup A_j\cup A_k\cup A_m|\\=4\cdot|A_i|-4\cdot3\cdot |A_i\cap A_j|+4!\cdot |A_i\cap A_j\cap A_k|-4!\cdot|A_i\cap A_j\cap A_k \cap A_m|\\=47520-116640+139968-46656=24192. $$
This result makes no sense as it exceeds the cardinality of the sample space. However, I don't see my mistake!?
EDIT:
Using the hints from the comments and redefining the set $A_i$ by "in column $i$ is exactly one rook", inclusion/exclusion principle yields: $$ |A_i\cup A_j\cup A_k\cup A_m|\\={4\choose 1}\cdot|A_i|-{4\choose 2}\cdot |A_i\cap A_j|+{4\choose 3}\cdot |A_i\cap A_j\cap A_k|-{4\choose 4}\cdot|A_i\cap A_j\cap A_k \cap A_m|\\=24192-19440+7776-1944=10584. $$
According to how the question is stated, the chess board is irrelevant, and the property of rooks are irrelevant. So the question simplifies massively. We have 4 "rooks", and 4 "columns", and each column has 3 "spots". Since each column must contain a rook, we must have exactly one rook in each column. There's $3^4$ possibilities if the rooks were indistinguishable, but the question says that the rooks are "different" (meaning "distinguishable"), so that's $4! \cdot 3^4$ possibilities.
For this problem the distinction between "distinguishable" and "indistinguishable" rooks is just a factor $4!$ throughout. I considered indistinguishable rooks.
Inclusion-exclusion is the wrong tool for the job here---you're making it about 1000 times harder than it needs to be. The first error in your proof is in counting $|A_i|$; it's clearly wrong since your value of $|A_i|$ is equal to $|\Omega|$. To count $|A_i|$ you'll need to account for the different ways column $i$ can be filled. There are $\binom{3}{k}$ ways to put $k$ rooks in column $i$, after which there are $[9]_{4-k}$ ways of placing the remaining rooks (where $[a]_b = a \times \cdots \times a-b+1$). Thus $$|A_i| = \sum_{k=1}^3 \binom{3}{k} [9]_{4-k}.$$ You'll then need to do even more elaborate enumeration for the other terms.
There's a good chance the actual question asks about non-attacking rooks, and you're meant to compute the number of derangements on 4 elements, for which Inclusion-Exclusion is a suitable tool.
If the "main diagonal" constraint didn't exist, there would be $4!$ ways of placing the four non-attacking rooks. For $S \subseteq \{1,2,3,4\}$, define $$A_S = \{\text{combinations in which the (non-attacking) rooks in columns } i \in S \text{ are all on the main diagonal}\}$$ and by Inclusion-Exclusion, there are $$ \sum_{S \subseteq \{1,2,3,4\}} (-1)^{|S|} |A_S| $$ ways to place the rooks so they don't intersect the main diagonal. (I like to think of $S$ as a "clash set": we insist on these clashes, but there may be others.)
From here, we (a) compute a formula for $|A_S|$, and (b) note that $|A_{S_1}|=|A_{S_2}|$ when $|S_1|=|S_2|$ which allows us to simplify the equation.