In Young's double slit experiment, we have two slits separated by a distance $$d$$, and a screen placed at a distance $$L$$ from the slits. When light with a single wavelength $$\lambda$$ passes through the slits, an interference pattern is formed on the screen with bright and dark fringes.

In this problem, we have a beam of light consisting of two wavelengths $$\lambda_1 = 7000~\mathop A\limits^o$$ and $$\lambda_2 = 5500~\mathop A\limits^o$$. We are given the distance between the slits $$d = 2.5 \mathrm{~mm}$$ and the distance between the plane of the slits and the screen $$L = 150 \mathrm{~cm}$$. Our goal is to find the least distance from the central fringe where the bright fringes due to both wavelengths coincide.

First, let's convert the wavelengths and distances to meters:

$$\lambda_1 = 7 \times 10^{-7} \mathrm{~m}$$ $$\lambda_2 = 5.5 \times 10^{-7} \mathrm{~m}$$ $$d = 2.5 \times 10^{-3} \mathrm{~m}$$ $$L = 1.5 \mathrm{~m}$$

The fringe width $$\beta$$ for a single wavelength is given by:

$$\beta = \frac{\lambda D}{d}$$

Now, let's consider the condition for the bright fringes due to both wavelengths to coincide. Let the $$n^{\text{th}}$$ bright fringe of $$\lambda_1$$ match with the $$m^{\text{th}}$$ bright fringe of $$\lambda_2$$. In this case, we have:

$$n \beta_1 = m \beta_2$$

Substituting the expression for fringe width, we get:

$$n \frac{\lambda_1 L}{d} = m \frac{\lambda_2 L}{d}$$

Simplifying, we get:

$$\frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{11}{14}$$

The least values of $$n$$ and $$m$$ that satisfy this condition are $$n = 11$$ and $$m = 14$$.

Now, let's find the position $$y$$ of the coincident bright fringe on the screen:

$$y = n \beta_1 = n \frac{\lambda_1 L}{d} = \frac{11 \times 7 \times 10^{-7} \mathrm{~m} \times 1.5 \mathrm{~m}}{2.5 \times 10^{-3} \mathrm{~m}}$$

$$y = k \times 10^{-5} \mathrm{~m}$$

Calculating the value of $$k$$:

$$k = \frac{11 \times 7 \times 10^{-7} \mathrm{~m} \times 1.5 \mathrm{~m}}{2.5 \times 10^{-3} \mathrm{~m} \times 10^{-5} \mathrm{~m}} = 462$$

So, the least distance from the central fringe where the bright fringes due to both wavelengths coincide is $$462 \times 10^{-5} \mathrm{~m}$$.