The problem:

Chloroform, $\ce{CHCl3}$, is formed from methane and chlorine in the following reaction:

$$\ce{CH4(g) + 3Cl2(g) ->3HCl(g) + CHCl3(g)}$$

Calculate $\Delta rH^{\circ}$, the enthalpy of formation for this reaction, using the enthalpies of formation of $\ce{CO2(g)}$ (-393.509kJ/mol-rxn), $\ce{H2O(l)}$ (-285.83kJ/mol-rxn) and $\ce{CHCl3(g)}$ (-103.1kJ/mol-rxn) and the enthalpy changes for the following reactions:

$$\ce{(1) \ \ CH4(g) + 2O2(g) -> 2H2O(l) + CO2(g) \ \ \ \ \Delta rH^{\circ}=-890.4kJ/mol-rxn \\ (2) \ \ 2HCl(g) -> H2(g) + Cl2(g) \ \ \ \ \Delta rH^{\circ}=+184.6kJ/mol-rxn }$$

My attempt at the problem:

Given the hint in the problem statement and by looking at the two given reaction equations i assumed that i needed to create another equation. I created an equation for $\ce{CHCl3(g)}$ from $\ce{CO2(g)}$ and $\ce{H2O(l)}$ and calculated it's $\Delta rH^{\circ}$ from the enthalpies of formation as shown below (Let's call this equation 3):

$$\ce{2CO2(g) + 5H2(g) + 3Cl2(g) -> 2CHCl3(g) + 4H2O(l)}$$

$\Delta rH^{\circ} = -562.502kJ/mol$

Here's where i run into difficulties: I can see that in the 'target reaction' we have HCl on the right and CHCl3 on the right, so it stands to reason that equation 3 will stand as it is presented but equation 2 will switch direction (and the sign of the enthalpy of reaction swop too). I don't want to switch the direction of equation 1 around because it has CH4 on the left and in the target reaction CH4 is on the left. We could multiply equation 1 by 2 and equation 2 by 3 in order to have equal amounts of chlorines and Co2's so we can cancel them, however we are now left with hydrogens on the left and H2O's on the right with no way to cancel them. Hydrogen and water obviously don't appear in the target reaction so i have gone wrong somewhere along the line.

Am i missing a reaction equation? Is there something i can do to cancel the unwanted reactants in the reaction equations? Any help would be greatly appreciated.