At 298 $\mathrm{K}$ the standard reduction potentials are 1.51 $\mathrm{V}$ for ${\mathrm{MnO}}_4^{-}\mid {\mathrm{Mn}}^{2+},1.36\mathrm{V}$ for ${\mathrm{Cl}}_2\mid {\mathrm{Cl}}^{-},1.07\mathrm{V}$ for ${\mathrm{Br}}_2\mid {\mathrm{Br}}^{-}$ and 0.54 V for ${\mathrm{I}}_2\mid {\mathrm{I}}^{-}\mathrm{AtpH}=3$ , permanganate is expected to oxidize : (2.303 RT/F = 0.059 V)

The reduction reaction is 

${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$

Its reduction potential is $E=E^{\circ }-\frac{RT}{5F}\ln \left(\frac{\left[{\mathrm{Mn}}^{2+}\right]}{\left[{\mathrm{MnO}}_4^{-}\right]{\left[{\mathrm{H}}^{+}\right]}^8}\right)$ Only numerical values of the terms within brackets are to be used. For $\mathrm{pH}= 3,$we have $\left[{\mathrm{H}}^{+}\right]={10}^{-3}\mathrm{M}$. With this, the above expression becomes $E=E^{\circ }-\frac{RT}{5F}\ln \left({10}^{24}\right)-\frac{RT}{5F}\ln \left(\frac{\left[{\mathrm{Mn}}^{2+}\right]}{\left[{\mathrm{MnO}}_4^{-}\right]}\right)$ 

Assuming$\left[{\mathrm{Mn}}^{2+}\right]=1\mathrm{M}$ and $\left[{\mathrm{MnO}}_4^{-}\right]=1\mathrm{M}$ we get $E=1.51\mathrm{V}-\left(\frac{0.059\mathrm{V}}{5}\right)24=1.51\mathrm{V}-0.283\mathrm{V}=1.227\mathrm{V}$

With this potential, only ${\mathrm{Br}}_2$ and ${\mathrm{I}}_2$ will be oxidized as their reduction potentials are smaller than 1.227 V. Chlorine will not be oxidized as its potential is larger than 1.227 V.