If $$3^{2\sin 2\alpha - 1}$$, 14 and $$3^{4 - 2\sin 2\alpha}$$ are the first three terms of an A.P. for some $$\alpha$$, then the sixth term of this A.P. is:

We are told that $$3^{2\sin 2\alpha - 1}$$, 14 and $$3^{4 - 2\sin 2\alpha}$$ are consecutive terms of an arithmetic progression (A.P.).

For any three consecutive terms $$A,\,B,\,C$$ in an A.P., the defining property is

$$2B \;=\; A + C.$$

Using this property with the given terms, we write

$$2 \times 14 \;=\; 3^{2\sin 2\alpha - 1} \;+\; 3^{4 - 2\sin 2\alpha}.$$

Simplifying the left side gives

$$28 \;=\; 3^{2\sin 2\alpha - 1} \;+\; 3^{4 - 2\sin 2\alpha}.$$

To handle the exponents cleanly, let us introduce

$$x \;=\; 2\sin 2\alpha - 1.$$

Then $$3^{2\sin 2\alpha - 1} = 3^{x}.$$ Now notice that

$$4 - 2\sin 2\alpha \;=\; 4 - (x + 1) \;=\; 3 - x,$$

so that

$$3^{4 - 2\sin 2\alpha} \;=\; 3^{3 - x}.$$

Substituting these into the equation, we obtain

$$3^{x} \;+\; 3^{3 - x} \;=\; 28.$$

Rewriting $$3^{3 - x}$$ as $$3^{3}\,3^{-x} = 27\,3^{-x},$$ we have

$$3^{x} + 27\,3^{-x} = 28.$$

Now set $$y = 3^{x}.$$ Because an exponential of base 3 is always positive, we know $$y > 0.$$ Also, $$3^{-x} = \dfrac{1}{3^{x}} = \dfrac{1}{y}.$$ Thus the equation becomes

$$y + 27\left(\dfrac{1}{y}\right) = 28.$$

Multiplying through by $$y$$ to clear the denominator, we arrive at the quadratic

$$y^{2} - 28y + 27 = 0.$$

We solve this using the quadratic formula. The discriminant is

$$\Delta = (-28)^{2} - 4 \times 1 \times 27 = 784 - 108 = 676,$$

and since $$\sqrt{676} = 26,$$ the two roots are

$$y = \frac{28 \pm 26}{2}.$$

Hence

$$y_{1} = \frac{28 + 26}{2} = 27, \quad y_{2} = \frac{28 - 26}{2} = 1.$$

Remembering that $$y = 3^{x},$$ we examine both possibilities:

1. $$y = 27 \implies 3^{x} = 3^{3} \implies x = 3.$$     Then $$2\sin 2\alpha - 1 = 3 \implies 2\sin 2\alpha = 4 \implies \sin 2\alpha = 2,$$ which is impossible because the sine of any angle lies between −1 and 1. Hence this root is rejected.

2. $$y = 1 \implies 3^{x} = 3^{0} \implies x = 0.$$     Therefore $$2\sin 2\alpha - 1 = 0 \implies 2\sin 2\alpha = 1 \implies \sin 2\alpha = \dfrac{1}{2},$$ which is perfectly admissible.

Thus the only valid solution is $$\sin 2\alpha = \dfrac{1}{2}.$$ Substituting this back to find the actual terms of the A.P.:

First term: $$3^{2\sin 2\alpha - 1} = 3^{2\left(\frac{1}{2}\right) - 1} = 3^{1 - 1} = 3^{0} = 1.$$

Second term: given as 14.

Third term: $$3^{4 - 2\sin 2\alpha} = 3^{4 - 1} = 3^{3} = 27.$$

We can quickly verify the common difference $$d$$:

$$d = 14 - 1 = 13,$$ $$27 - 14 = 13,$$ so $$d = 13.$$

The general formula for the $$n^{\text{th}}$$ term of an A.P. is

$$a_{n} = a_{1} + (n - 1)d.$$

Taking $$n = 6$$, we have

$$a_{6} = 1 + (6 - 1)\times 13 = 1 + 5 \times 13 = 1 + 65 = 66.$$

Hence, the correct answer is Option A.