If $$\frac{3^{a + 3} \times 4^{a + 6} \times 25^{a + 1}}{27^{a - 1} \times 8^{a - 2} \times 125^{a + 4}} = \frac{4}{15^{26}}$$, then the value of $$\sqrt{a + 9}$$ is:

$$\frac{3^{a+3}\times4^{a+6}\times25^{a+1}}{27^{a-1}\times8^{a-2}\times125^{a+4}}=\frac{4}{15^{26}}$$

$$\frac{3^{a+3}\times2^{2\left(a+6\right)}\times5^{2\left(a+1\right)}}{3^{3\left(a-1\right)}\times2^{3\left(a-2\right)}\times5^{3\left(a+4\right)}}=\frac{2^2}{3^{26}\times5^{26}}$$

$$\frac{3^{a+3}\times2^{2a+12}\times5^{2a+2}}{3^{3a-3}\times2^{3a-6}\times5^{3a+12}}=\frac{2^2}{3^{26}\times5^{26}}$$

$$\frac{2^{2a+12-3a+6}}{3^{3a-3-a-3}\times5^{3a+12-2a-2}}=\frac{2^2}{3^{26}\times5^{26}}$$

$$\frac{2^{-a+18}}{3^{2a-6}\times5^{a+10}}=\frac{2^2}{3^{26}\times5^{26}}$$

Comparing both sides

-a + 18 = 2

$$=$$>  a = 16

$$\therefore\ $$ $$\sqrt{a + 9}$$ = $$\sqrt{16+9}$$ = $$\sqrt{25}$$ = 5

Hence, the correct answer is Option A