Calculating the dissociation constant of one acid in a mixture
I just wanted to add an answer based on "no" assumptions, except (and that is occasionally a big assumption) that concentration equals activity. At first it does not seem very practical, as of its size, but I couldn't stop me from adding it here as I like to see the whole picture. In the end, it is actually not that hard.

The exact equation for the proton concentration is:

$$[\ce{H+}] = [\ce{OH-}] + [\ce{Cl-}] + [\ce{CHCl2COO-}] + [\ce{CH3COO-}]$$

which is the sum of the hydroxide ions from the autodissociation of water and the sum of the resulting acid residue ions from the three acids. As is shown here [1], this system expands to the following (and can be reduced again, if appropriate simplifications are taken into account for each acid, as is shown here [2]):

$$
[\ce{H+}] = [\ce{OH-}]
+ \frac{[\ce{HCl}]_0 K_{\mathrm a1}}{[\ce{H+}] + K_{\mathrm a1}}
+ \frac{[\ce{CHCl2COOH}]_0 K_{\mathrm a2}}{[\ce{H+}] + K_{\mathrm a2}}
+ \frac{[\ce{CH3COOH}]_0 K_{\mathrm a3}}{[\ce{H+}] + K_{\mathrm a3}}
$$

where $K_{\mathrm a1}$, $K_{\mathrm a2}$, and $K_{\mathrm a3}$ are the $K_{\mathrm a}$'s of $\ce{HCl}$, $\ce{CHCl2COOH}$, and $\ce{CH3COOH}$ respectively.

Now what you need to find is $K_{\mathrm a2}$. Solving it at this stage for $K_{\mathrm a2}$ is probably not a good idea aka nothing nice to handle, as the system is quite large. But as you know all values except $K_{\mathrm a2}$, it's actually not that bad.

$$
\underbrace{[\ce{H+}]}_\pu{0.1 mol/l} = \underbrace{[\ce{OH-}]}_\pu{10^{-13} mol/l}
+ \underbrace{\frac{[\ce{HCl}]_0 K_{\mathrm a1}}{[\ce{H+}] + K_{\mathrm a1}}}_\pu{0.09 mol/l}
+ \frac{[\ce{CHCl2COOH}]_0 K_{\mathrm a2}}{[\ce{H+}] + K_{\mathrm a2}}
+ \underbrace{\frac{[\ce{CH3COOH}]_0 K_{\mathrm a3}}{[\ce{H+}] + K_{\mathrm a3}}}_\pu{10^{-5} mol/l}
$$

$\ldots$ I'll leave units from here on (everything is in $\pu{mol/l}$). This simplifies quite much into:
$$\begin{align}
0.1 &amp;= 10^{-13} + 0.09 + \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}} + 10^{-5}\\
0.1 - 10^{-13} - 0.09 - 10^{-5} &amp;= \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}}\\
0.01^* &amp;= \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}}\\
0.01\,(0.1 + K_{\mathrm a2}) &amp;= 0.09\,K_{\mathrm a2}\\
0.001 + 0.01 K_{\mathrm a2} &amp;= 0.09 K_{\mathrm a2}\\
0.001 &amp;= 0.09 K_{\mathrm a2} - 0.01 K_{\mathrm a2}\\
0.001 &amp;= (0.09 - 0.01)\,K_{\mathrm a2}\\
0.001 &amp;= 0.08\,K_{\mathrm a2}\\
K_{\mathrm a2} &amp;= \frac{0.001}{0.08}\\
&amp;= 0.0125
\end{align}$$



* I rounded to $0.01$; it was actually $0.00999\ldots$ which would lead to $\pu{0.0124861... mol/l}$

[1] Equations 2.1 and 2.2
[2] Assumption 1 for HCl, Assumption 3 for acetic acid