$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

-or $\quad\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}} ;\left[\mathrm{H}^{+}\right]$ of solution $1=10^{-3}$

$\left[\mathrm{H}^{+}\right]$ of solution $2=10^{-4} ;\left[\mathrm{H}^{+}\right]$ of solution $3=10^{-5}$

Total concentration of $\left[\mathrm{H}^{+}\right]$

$=10^{-3}\left(1+1 \times 10^{-1}+1 \times 10^{-2}\right)$

$\Rightarrow 10^{-3}\left(\frac{1}{1}+\frac{1}{10}+\frac{1}{100}\right)$

$\Rightarrow 10^{-3}\left(\frac{100+10+1}{100}\right)$

$\Longrightarrow 10^{-3}\left(\frac{111}{100}\right)=1.11 \times 10^{-3}$

$So, \mathrm{H}^{+}$ ion concentration in mixture of equal volume of these acid solution $=1.11 \times 10^{-3} / 3=3.7 \times 10^{-4} \,\mathrm{M}$