Let $$I(x)=\int\frac{3dx}{\left(4x+6\right)\left(\sqrt{4x^{2}}+8x+3\right)}$$ and $$I(0)=\frac{{\sqrt{3}}}{4}+20.$$
If $$I\left( \frac{1}{2} \right)=\frac{a\sqrt{2}}{b}+c, \text { Where a,b,c } \in N,gcd(a,b)=1, \text{ a+b+c is equal to}$$

We need to evaluate $$I(x) = \int \frac{3\,dx}{(4x+6)\sqrt{4x^2+8x+3}}$$.

First, simplify the expression under the square root:

$$4x^2 + 8x + 3 = 4(x^2 + 2x) + 3 = 4(x+1)^2 - 4 + 3 = 4(x+1)^2 - 1$$

Also, $$4x + 6 = 2(2x + 3)$$.

So the integral becomes:

$$I(x) = \int \frac{3\,dx}{2(2x+3)\sqrt{4(x+1)^2 - 1}}$$

Let us try the substitution $$t = \frac{1}{2x+3}$$. Then $$2x + 3 = \frac{1}{t}$$, so $$x = \frac{1-3t}{2t}$$, and $$dx = -\frac{1}{2t^2}dt$$.

Now compute $$4x^2 + 8x + 3$$:

$$4x^2 + 8x + 3 = (2x+3)(2x+1)$$

We have $$2x + 3 = \frac{1}{t}$$, and $$2x + 1 = \frac{1}{t} - 2 = \frac{1-2t}{t}$$.

So $$4x^2 + 8x + 3 = \frac{1}{t} \cdot \frac{1-2t}{t} = \frac{1-2t}{t^2}$$.

$$\sqrt{4x^2 + 8x + 3} = \frac{\sqrt{1-2t}}{t}$$ (taking $$t > 0$$ for now).

Substituting into the integral:

$$I = \int \frac{3}{2 \cdot \frac{1}{t} \cdot \frac{\sqrt{1-2t}}{t}} \cdot \left(-\frac{1}{2t^2}\right) dt$$

$$= \int \frac{3t^2}{2\sqrt{1-2t}} \cdot \left(-\frac{1}{2t^2}\right) dt$$

$$= \int \frac{-3}{4\sqrt{1-2t}} dt$$

$$= \frac{-3}{4} \cdot \frac{\sqrt{1-2t}}{-1} + C = \frac{3}{4}\sqrt{1-2t} + C$$

Substituting back $$t = \frac{1}{2x+3}$$:

$$I(x) = \frac{3}{4}\sqrt{1 - \frac{2}{2x+3}} + C = \frac{3}{4}\sqrt{\frac{2x+3-2}{2x+3}} + C = \frac{3}{4}\sqrt{\frac{2x+1}{2x+3}} + C$$

Now use the initial condition $$I(0) = \frac{\sqrt{3}}{4} + 20$$:

$$I(0) = \frac{3}{4}\sqrt{\frac{1}{3}} + C = \frac{3}{4} \cdot \frac{1}{\sqrt{3}} + C = \frac{\sqrt{3}}{4} + C$$

So $$C = 20$$.

Therefore $$I(x) = \frac{3}{4}\sqrt{\frac{2x+1}{2x+3}} + 20$$.

Now compute $$I\left(\frac{1}{2}\right)$$:

$$I\left(\frac{1}{2}\right) = \frac{3}{4}\sqrt{\frac{2 \cdot \frac{1}{2} + 1}{2 \cdot \frac{1}{2} + 3}} + 20 = \frac{3}{4}\sqrt{\frac{2}{4}} + 20 = \frac{3}{4} \cdot \frac{\sqrt{2}}{2} + 20 = \frac{3\sqrt{2}}{8} + 20$$

So $$I\left(\frac{1}{2}\right) = \frac{3\sqrt{2}}{8} + 20$$.

Comparing with $$\frac{a\sqrt{2}}{b} + c$$: we get $$a = 3$$, $$b = 8$$, $$c = 20$$.

Check: $$\gcd(3, 8) = 1$$ ✓, and $$a, b, c \in \mathbb{N}$$ ✓.

$$a + b + c = 3 + 8 + 20 = 31$$

The answer is Option B: $$31$$.