In a hydrogen spectrum $$\lambda$$ be the wavelength of first transition line of Lyman series. The wavelength difference will be "$$a\lambda$$" between the wavelength of $$3^{rd}$$ transition line of Paschen series and that of $$2^{nd}$$ transition line of Balmer Series where $$a =$$ ______.

We need to find the wavelength difference between the 3rd transition line of the Paschen series and the 2nd transition line of the Balmer series in terms of $$\lambda$$, the wavelength of the first transition line of the Lyman series.

We start by finding the wavelength of the first transition line of the Lyman series ($$n = 2 \to n = 1$$):

$$\dfrac{1}{\lambda} = R\left(\dfrac{1}{1^2} - \dfrac{1}{2^2}\right) = R\left(1 - \dfrac{1}{4}\right) = \dfrac{3R}{4}$$

This gives $$\lambda = \dfrac{4}{3R}$$.

Next, for the 3rd transition line of the Paschen series ($$n = 6 \to n = 3$$), we have:

$$\dfrac{1}{\lambda_1} = R\left(\dfrac{1}{3^2} - \dfrac{1}{6^2}\right) = R\left(\dfrac{1}{9} - \dfrac{1}{36}\right) = R\left(\dfrac{4 - 1}{36}\right) = \dfrac{R}{12}$$

Therefore, $$\lambda_1 = \dfrac{12}{R}$$.

Now, considering the 2nd transition line of the Balmer series ($$n = 4 \to n = 2$$):

$$\dfrac{1}{\lambda_2} = R\left(\dfrac{1}{2^2} - \dfrac{1}{4^2}\right) = R\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = R\left(\dfrac{4 - 1}{16}\right) = \dfrac{3R}{16}$$

This gives $$\lambda_2 = \dfrac{16}{3R}$$.

Calculating the difference yields:

$$\lambda_1 - \lambda_2 = \dfrac{12}{R} - \dfrac{16}{3R} = \dfrac{36 - 16}{3R} = \dfrac{20}{3R}$$

Since $$\lambda = \dfrac{4}{3R}$$, it follows that $$\dfrac{1}{R} = \dfrac{3\lambda}{4}$$, and hence:

$$\lambda_1 - \lambda_2 = \dfrac{20}{3R} = \dfrac{20}{3} \times \dfrac{3\lambda}{4} = \dfrac{20\lambda}{4} = 5\lambda$$

Therefore, $$a = \boxed{5}$$.

The value of $$a$$ is 5.