Let $$z$$ be a complex number such that $$|z + 2| = 1$$ and $$\text{Im}\left(\frac{z+1}{z+2}\right) = \frac{1}{5}$$. Then the value of $$|\text{Re}(z + 2)|$$ is

Given $$|z + 2| = 1$$ and $$\text{Im}\left(\frac{z+1}{z+2}\right) = \frac{1}{5}$$, we wish to find $$|\text{Re}(z+2)|$$.

Let $$w = z + 2$$ so that $$|w| = 1$$. Writing $$w = a + bi$$ gives the relation $$a^2 + b^2 = 1$$ and hence $$z + 1 = w - 1 = (a-1) + bi$$.

Now, $$\frac{z+1}{z+2} = \frac{w-1}{w}$$. Since $$\frac{w-1}{w} = 1 - \frac{1}{w} = 1 - \frac{\bar{w}}{|w|^2} = 1 - \bar{w} = 1 - (a - bi) = (1-a) + bi$$, it follows that $$\text{Im}\left(\frac{z+1}{z+2}\right) = b = \frac{1}{5}$$.

Using the unit-circle condition $$a^2 + b^2 = 1$$ with $$b = \frac{1}{5}$$ yields $$a^2 + \frac{1}{25} = 1 \implies a^2 = \frac{24}{25} \implies |a| = \frac{2\sqrt{6}}{5}$$.

Since $$\text{Re}(z+2) = a$$, we conclude that $$|\text{Re}(z+2)| = |a| = \frac{2\sqrt{6}}{5}$$.

The correct answer is Option (1): $$\frac{2\sqrt{6}}{5}$$.