Multiple solutions in the balancing of redox reactions
The first thing that you should realise is that under specific conditions (here: acidic ones) only one product is formed in this reaction. Permanganate is not reduced to both manganese(IV) oxide and manganese(II) cations, it will either cleanly give manganese(IV) oxide (if the acid concentration is very low) or cleanly give manganese(II) cations (if the acid concentration is sufficient). The only way to generate both is to have sufficient acid at first which is subsequently used up by the reaction and drops below a threshold. Thus, if you were titrating and notice the dark precipitate of $\ce{MnO2}$, immediately stop, throw away the analysis and start again as it has become worthless.

Therefore, there is only one simplest solution to the problem, depending on which half-reaction you prefer:

$$\begin{align}\ce{MnO4- + 8 H+ + 5 e- &amp;-&gt; Mn^2+ + 4 H2O}\tag{Red1}\\[0.6em]
\ce{H2O2 &amp;-&gt; O2 + 2 H+ + 2e-}\tag{Ox}\\[0.6em]
\ce{2 MnO4- + 6 H+ + 5 H2O2 &amp;-&gt; 2 Mn^2+ + 8 H2O + 5 O2}\tag{Redox1}\\[1.6em]
\ce{MnO4- + 3 e- + 4 H+ &amp;-&gt; MnO2 + 2 H2O}\tag{Red2}\\[0.6em]
\ce{H2O2 &amp;-&gt; O2 + 2 H+ + 2 e-}\tag{Ox}\\[0.6em]
\ce{2 MnO4- + 2 H+ + 3 H2O2 &amp;-&gt; 2 MnO2 + 4 H2O}\tag{Redox2}\end{align}$$