If $$9(a^2 + b^2) + c^2 + 20 = 12(a + 2b)$$, then the value of $$\sqrt{6a + 9b + 2c}$$ is:

$$9(a^2+b^2)+c^2+20=12(a+2b)$$

$$9a^2+9b^2+c^2+20=12a+24b$$

$$9a^2-12a+9b^2-24b+c^2+20=0$$

$$9a^2-12a+4-4+9b^2-24b+16-16+c^2+20=0$$

$$\left(3a-2\right)^2-4+\left(3b-4\right)^2-16+c^2+20=0$$

$$\left(3a-2\right)^2+\left(3b-4\right)^2+c^2=0$$

$$3a-2=0,\ 3b-4=0,\ c=0$$

$$a=\frac{2}{3},\ b=\frac{4}{3},\ c=0$$

$$\sqrt{6a+9b+2c}=\sqrt{6\left(\frac{2}{3}\right)+9\left(\frac{4}{3}\right)+2\left(0\right)}$$

= $$\sqrt{4+12}$$

= $$\sqrt{16}$$

= 4

Hence, the correct answer is Option A