Let the $n^{\text{th}}$ bright fringe of wavelength $\lambda_1$ coincide with the $m^{\text{th}}$ bright fringe of wavelength $\lambda_2$. The position of the $n^{\text{th}}$ bright fringe in Young's double slit experiment is given by $y = \frac{n\lambda D}{d}$. Equating the positions for both wavelengths: $\frac{n\lambda_1 D}{d} = \frac{m\lambda_2 D}{d}$ $\implies n\lambda_1 = m\lambda_2$ $\implies \frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{10000\text{ \AA}}{12000\text{ \AA}} = \frac{5}{6}$ For the minimum distance from the central maximum, we take the smallest integers satisfying this ratio, which are $n = 5$ and $m = 6$. The minimum distance $y$ is: $y = \frac{n\lambda_1 D}{d} = \frac{5 \times 12000 \times 10^{-10}\text{ m} \times 2\text{ m}}{2 \times 10^{-3}\text{ m}}$ $y = 5 \times 12000 \times 10^{-7}\text{ m} = 60000 \times 10^{-7}\text{ m} = 6 \times 10^{-3}\text{ m} = 6\text{ mm}$