1. Let the first three terms of the A.P. be $a_1=3^{2\sin 2\alpha-1},\quad a_2=14,\quad a_3=3^{4-2\sin 2\alpha}.$

2. Since these are in A.P., the middle term is the arithmetic mean of the other two: $2a_2=a_1+a_3.$ So, $28=3^{2\sin 2\alpha-1}+3^{4-2\sin 2\alpha}.$

3. Put $x=2\sin 2\alpha-1.$ Then $4-2\sin 2\alpha=3-x,$ because $x=2\sin 2\alpha-1 \implies 2\sin 2\alpha=x+1,$ so $4-2\sin 2\alpha=4-(x+1)=3-x.$

Thus the equation becomes $3^x+3^{3-x}=28.$

4. Let $t=3^x.$ Then $3^{3-x}=\frac{27}{3^x}=\frac{27}{t}.$ So, $t+\frac{27}{t}=28.$ Multiplying by $t$, $t^2-28t+27=0.$

5. Solve the quadratic: $t^2-28t+27=0$ $=(t-1)(t-27)=0.$ Hence, $t=1 \quad \text{or} \quad t=27.$

6. Therefore,

• if $t=1$, then $a_1=1$ and $a_3=27$;
• if $t=27$, then $a_1=27$ and $a_3=1$.

Since the given order is first, second, third term of an A.P., we need $2\cdot 14=a_1+a_3,$ and the common difference must be consistent with the order.

Check:

• For $a_1=1, a_2=14, a_3=27$, the common difference is $13$, so this is a valid A.P.
• For $a_1=27, a_2=14, a_3=1$, the common difference is $-13$, which is also an A.P.

Now compute the sixth term in the valid increasing A.P. corresponding to the options: $a_6=a_1+5d=1+5(13)=66.$

7. Hence the required sixth term is $\boxed{66}.$

8. Comparison with stored answer: Stored correct answer is A, i.e. $66$, which matches our result.