Let the G.P. be

$a, ar, ar^2, ar^3, \dots$

where $a>0$ and $r>0$.

Use the given sum of 2nd, 4th and 6th terms

The 2nd, 4th and 6th terms are: $ar,\ ar^3,\ ar^5$

So, $ar+ar^3+ar^5=21$ $ar(1+r^2+r^4)=21 \quad ...(1)$

Use the given sum of 8th, 10th and 12th terms

The 8th, 10th and 12th terms are: $ar^7,\ ar^9,\ ar^{11}$

So, $ar^7+ar^9+ar^{11}=15309$ $ar^7(1+r^2+r^4)=15309 \quad ...(2)$

Divide (2) by (1)

$\frac{ar^7(1+r^2+r^4)}{ar(1+r^2+r^4)}=\frac{15309}{21}$

$r^6=729=3^6$

Since terms are positive, $r>0$, hence $r=3$

Find the first term $a$

From (1): $a(3)(1+9+81)=21$ $3a(91)=21$ $273a=21$ $a=\frac{21}{273}=\frac{1}{13}$

Find the sum of first 9 terms

$S_9=a\frac{r^9-1}{r-1}$

Substitute $a=\frac{1}{13}$ and $r=3$: $S_9=\frac{1}{13}\cdot \frac{3^9-1}{3-1}$

$3^9=19683$

$S_9=\frac{1}{13}\cdot \frac{19683-1}{2}$ $S_9=\frac{1}{13}\cdot \frac{19682}{2}$ $S_9=\frac{9841}{13}$

Since $13\times 757=9841$, $S_9=757$

Check options

The correct option is: $\boxed{\text{A: }757}$