Let $$a_{1},a_{2},a_{3},...$$ be a G.P. of increasing positive terms such that $$a_{2}.a_{3}.a_{4}=64\text{ and }a_{1}+a_{3}+a_{5}=\frac{813}{7}.\text{ Then }a_{3}+a_{5}+a_{7}$$ is equal to :

GP with increasing positive terms. $$a_2 \cdot a_3 \cdot a_4 = 64$$.

$$a_2 a_3 a_4 = (a_3/r)(a_3)(a_3 r) = a_3^3 = 64 \Rightarrow a_3 = 4$$.

$$a_1 + a_3 + a_5 = \frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$$

Let $$u = r^2$$: $$\frac{4}{u} + 4 + 4u = \frac{813}{7}$$

$$4u^2 + 4u + 4 = \frac{813u}{7}$$

$$28u^2 + 28u + 28 = 813u$$

$$28u^2 - 785u + 28 = 0$$

$$u = \frac{785 \pm \sqrt{785^2 - 4(28)(28)}}{56} = \frac{785 \pm \sqrt{616225 - 3136}}{56} = \frac{785 \pm \sqrt{613089}}{56} = \frac{785 \pm 783}{56}$$

$$u = \frac{1568}{56} = 28$$ or $$u = \frac{2}{56} = \frac{1}{28}$$.

Since GP is increasing: $$r > 1$$, so $$u = r^2 = 28$$.

$$a_3 + a_5 + a_7 = 4 + 4(28) + 4(28^2) = 4 + 112 + 3136 = 3252$$.

The answer is Option 4: 3252.