What is the net ionic equation of the following?
You are making a mistake splitting up your reaction equation from \eqref{full} to \eqref{wrong} in order to filter out the spectator ions.

\begin{align}
\ce{3Cl2(aq) + 2AlBr3(aq) &amp;-&gt; 2AlCl3(aq) + 3Br2(l)}\tag1\label{full}\\
\ce{6Cl⁻(aq) + 2Al^{3+}(aq) + 6Br-(aq) &amp;-&gt; 2Al^{3+}(aq) + 6Cl-(aq) + 3Br2(l)}\tag{wrong}\label{wrong}
\end{align}

What you are essentially doing is not conserving the charges. Dichlorine $\ce{Cl2}$ is a neutral molecule, and the following is what your transition implies, but is also wrong:
$$\ce{Cl2 -&gt; 2Cl-}\tag{also wrong}$$

The correct way to transform \eqref{full} is to split (dissociate) only the ionic compounds $\ce{AlBr3}$ and \ce{AlCl3}, and then cancel the spectator ions:

\begin{align}
\ce{3 Cl2 (aq) + 2 AlBr3 (aq) &amp;-&gt; 2 AlCl3 (aq) + 3 Br2 (l)}\tag1\\
\ce{3 Cl2 (aq) + 2 Al^3+ (aq) + 6 Br- (aq) &amp;-&gt; 
    2 Al^3+ (aq) + 6 Cl- (aq) + 3 Br2 (l)}\tag2\label{split}\\
\end{align}

Next step is to cancel the aluminium ions on each side, and reduce the equation to \eqref{net-ionic}:

\begin{align}
\ce{3 Cl2 (aq) + 6 Br- (aq) &amp;-&gt; 6 Cl- (aq) + 3 Br2 (l)}\tag3\\
\ce{Cl2 (aq) + 2 Br- (aq) &amp;-&gt; 2 Cl- (aq) + Br2 (l)}\tag4\label{net-ionic}
\end{align}

I don't think the reaction will actually produce liquid bromine, but aqueous bromine instead; that doesn't matter to the solution though.