Equations for the oxidation reactions of tea teaflavins and EGCG with potassium permanganate
The two organic compounds are ultimately being converted to carbon dioxide, $\ce{CO2}$, and water, $\ce{H2O}$, just as they would be in complete combustion with pure oxygen gas, $\ce{O2}$, in a bomb calorimeter. Thus the two balanced combustion equations would be as follows:

$$\ce{C29H24O12 + 29 O2 -&gt; 29 CO2 + 12 H2O} \tag{1}$$

$$\ce{C22H18O11 + 21 O2 -&gt; 22 CO2 + 9 H2O} \tag{2}$$

So now assume that potassium permanganate, $\ce{KMnO4}$, is being used as the oxidizing agent, with manganese dioxide, $\ce{MnO2}$, as the product. Then, as per Ron's answer (https://chemistry.stackexchange.com/a/85865/79678), and assumed reasonable conditions therein, we have the following reaction:

$$\ce{4 MnO4- + 4 H+ -&gt; 4 MnO2 + 3 O2 + 2 H2O} \tag{3}$$

Showing the spectator potassium ions, and arbitrarily assuming hydrochloric acid as the source of the hydrogen ions, yields

$$\ce{4 KMnO4 + 4 HCl -&gt; 4 MnO2 + 3 O2 + 2 H2O + 4 KCl} \tag{4}$$

Since 29 $\ce{O2}$ and 3 $\ce{O2}$ have no common factor other than unity, multiply equation 1 by 3, equation 4 by 29, add them, cancel the equal number of $\ce{O2}$ molecules on each side and the result is

$$\ce{3 C29H24O12 + 116 KMnO4 + 116 HCl -&gt; 87 CO2 + 116 MnO2 + 94 H2O + 116 KCl} \tag{5}$$

Since 21 $\ce{O2}$ is 7 times 3 $\ce{O2}$, use equation 2 as is, multiply equation 4 by 7, add them, cancel the equal number of $\ce{O2}$ molecules on each side and the result is

$$\ce{C22H18O11 + 28 KMnO4 + 28 HCl -&gt; 22 CO2 + 28 MnO2 + 23 H2O + 28 KCl} \tag{6}$$