$500 \mathrm{ml}$ of $2{\mathrm{MCH}}_3\mathrm{COOH}$ solution is mixed with $600 \mathrm{ml} 12\mathrm{\%}\mathrm{w}/{\mathrm{vCH}}_3\mathrm{COOH}$ solution then calculate the final molarity of solution.

$12 \mathrm{gm} {\mathrm{CH}}_3\mathrm{COOH}$ is present in 100 ml of solution

$120 \mathrm{gm} {\mathrm{CH}}_3\mathrm{COOH}$ is present in 1000 ml of solution

$M_2=\frac{120}{60}=2$ Now we are mixing 

$\begin{array}{l}500\mathrm{ml},2{\mathrm{CH}}_3\mathrm{COOH}+2M,600 \mathrm{ml} {\mathrm{CH}}_3\mathrm{COOH}\\ M_1{V}_1+M_2{V}_2=M_3{V}_3\\ 500\times 2+600\times 2=M_3\times 1100\\ M_3=\frac{2200}{1100}=2\end{array}$