Convert $A⟺B⟺C$ into Conjunctive Normal Form. 

*

*Eliminate implications and bi-conditionals using formulas:
$A⟺B⟺C$

Attempt:

$≡ (A⟹B)∧(B⟹A)⟺C$


$≡ (((A⟹B)∧(B⟹A))⟹C)∧(C⟹((A⟹B)∧(B⟹A)))$


$≡ ¬((A⟹B)∧(B⟹A))∨C ∧ ¬C∨((A⟹B)∧(B⟹A))$


$≡ ¬((¬A∨B)∧(¬B∨A))∨C ∧ ¬C∨((¬A∨B)∧(¬B∨A))$




*Apply De-Morgan’s Law and reduce NOT symbols so as to bring negations before the atoms. Use:


Attempt:

$≡ ¬((¬A∨B)∧(¬B∨A))∨C ∧ ¬C∨((¬A∨B)∧(¬B∨A))$


$≡ ¬((¬A∨B)∧(¬B∨A))∨C ∧ ¬(C∨(¬A∨B)∧(¬B∨A))$


$≡ ((¬¬A∨¬B)∧(¬¬B∨¬A)∨C ∧(¬C∨¬¬A∨B)∧(¬C∨¬¬B∨A)$


$≡ (A∨¬B∨C)∧(B∨¬A∨C)∧(¬C∨A∨B)∧(¬C∨B∨A) == \text{CNF}$

 A: The first part of your attempt is correct, but there are some mistakes in the second part.
Here is your second part:
\begin{align}
& (¬((¬A∨B)∧(¬B∨A))∨C) ∧ (\color{red}¬C∨(¬A∨B)∧(¬B∨A))\tag{1}\\
≡& (\color{lightblue}¬((¬A∨B)∧(¬B∨A))∨C) ∧ \color{red}¬(C∨(¬A∨B)∧(¬B∨A))\tag{2}\\
≡& ((\color{lightblue}¬¬A∨\color{lightblue}¬B)∧(\color{lightblue}¬¬B∨\color{lightblue}¬A)∨C) ∧(\color{red}¬C∨\color{red}¬¬A∨B)∧(\color{red}¬C∨\color{red}¬¬B∨A)\tag{3}\\
≡& (A∨¬B∨C)∧(B∨¬A∨C)∧(¬C∨A∨B)∧(¬C∨B∨A) \tag{CNF}\\
\end{align}
From step $(1)$ to step $(2)$, that just messed up the brackets, it's the negation of $C$,  not the whole thing.
From step $(2)$ to step $(3)$, we can't distribute the negation into the 
brackets without change any logic connectives. Apply De-Morgan’s Law correctly we should get:
\begin{align}
& (\color{lightblue}¬((¬A∨B)∧(¬B∨A))∨C) ∧ \color{red}¬(C∨(¬A∨B)∧(¬B∨A))\tag{2}\\
≡& ((\color{lightblue}¬¬A\land \color{lightblue}¬B)\lor(\color{lightblue}¬¬B\land\color{lightblue}¬A)\lor C) ∧(\color{red}¬C\land((\color{red}¬¬A\land \color{red}\neg B)\lor(\color{red}¬¬B\land\color{red}\neg A)))\\
\end{align}
However, since there already is a mistake in your previous step, so we need to go back and start over again.
\begin{align}
& (¬((¬A∨B)∧(¬B∨A))∨C) ∧ (¬C∨(¬A∨B)∧(¬B∨A))\tag{1}\\
\equiv&((A\land\neg B)\lor(B\land\neg A)\lor C) ∧ (¬C∨(¬A∨B)∧(¬B∨A))\\
\equiv&((A\land\neg B)\lor(B\lor C)\land(\neg A\lor C)) ∧ (¬C∨(¬A∨B)∧(¬B∨A))\\
\equiv&(((A\land\neg B)\lor B\lor C)\land((A\land\neg B)\lor\neg A\lor C)) ∧ (¬C∨(¬A∨B)∧(¬B∨A))\\
\equiv&(((A\lor B)\land(\neg B\lor B))\lor C)\land((A\land\neg B)\lor\neg A\lor C) ∧ (¬C∨(¬A∨B)∧(¬B∨A))\\
\equiv&(((A\lor B)\land\top)\lor C)\land((A\land\neg B)\lor\neg A\lor C) ∧ (¬C∨(¬A∨B)∧(¬B∨A))\\
\equiv&(A\lor B\lor C)\land((A\land\neg B)\lor\neg A\lor C) ∧ (¬C∨(¬A∨B)∧(¬B∨A))\\
\equiv&(A\lor B\lor C)\land(((A\lor \neg A)\land(\neg B\lor \neg A))\lor C) ∧ (¬C∨(¬A∨B)∧(¬B∨A))\\
\equiv&(A\lor B\lor C)\land((\top\land(\neg B\lor \neg A))\lor C) ∧ (¬C∨(¬A∨B)∧(¬B∨A))\\
\equiv&(A\lor B\lor C)\land(\neg B\lor \neg A\lor C) ∧ (¬C∨(¬A∨B)∧(¬B∨A))\\
\equiv&(A\lor B\lor C)\land(\neg B\lor \neg A\lor C) ∧ (¬C∨¬A∨B)∧(¬C∨¬B∨A)\tag{CNF}\\
\end{align}
