If $$z_1, z_2$$ are complex numbers such that $$Re(z_1) = |z_1 - 1|$$ and $$Re(z_2) = |z_2 - 1|$$ and $$\arg(z_1 - z_2) = \frac{\pi}{6}$$, then $$Im(z_1 + z_2)$$ is equal to:

Let us denote $$z_1 = x_1 + i\,y_1$$ and $$z_2 = x_2 + i\,y_2$$, where $$x_1,x_2$$ are the real parts and $$y_1,y_2$$ are the imaginary parts.

For every complex number $$z$$ we have the identity $$Re(z)=x$$ when $$z=x+iy$$. The condition $$Re(z)=|z-1|$$ therefore becomes

$$x = |(x + iy) - 1| = |\, (x-1) + iy \,|.$$

The modulus formula $$|a+ib|=\sqrt{a^{2}+b^{2}}$$ gives

$$x = \sqrt{(x-1)^{2}+y^{2}}.$$

Squaring both sides,

$$x^{2} = (x-1)^{2} + y^{2}.$$

Expanding the right-hand side,

$$x^{2} = x^{2} - 2x + 1 + y^{2}.$$

Subtracting $$x^{2}$$ from both sides yields

$$0 = -2x + 1 + y^{2}.$$

Rearranging, we obtain the relation

$$y^{2} = 2x - 1.$$

This equation must be satisfied by both $$z_1$$ and $$z_2$$, so we have

$$y_1^{2} = 2x_1 - 1 \quad\text{and}\quad y_2^{2} = 2x_2 - 1.$$

Solve each for the real part:

$$x_1 = \frac{y_1^{2}+1}{2}, \qquad x_2 = \frac{y_2^{2}+1}{2}.$$

The argument condition $$\arg(z_1 - z_2)=\dfrac{\pi}{6}$$ tells us that the line segment from $$z_2$$ to $$z_1$$ makes an angle $$\dfrac{\pi}{6}$$ with the positive real axis. Hence the slope of that segment equals $$\tan\!\left(\dfrac{\pi}{6}\right)=\dfrac{1}{\sqrt{3}}$$, so

$$\frac{y_1 - y_2}{\,x_1 - x_2\,} = \frac{1}{\sqrt{3}}.$$

Substituting $$x_1$$ and $$x_2$$ obtained above,

$$\frac{y_1 - y_2}{\dfrac{y_1^{2}+1}{2} - \dfrac{y_2^{2}+1}{2}} = \frac{y_1 - y_2}{\dfrac{y_1^{2}-y_2^{2}}{2}} = \frac{y_1 - y_2}{\dfrac{(y_1-y_2)(y_1+y_2)}{2}} = \frac{2}{y_1 + y_2} = \frac{1}{\sqrt{3}}.$$

Multiplying both sides by $$y_1+y_2$$ and by $$\sqrt{3}$$ gives

$$2 = \frac{y_1 + y_2}{\sqrt{3}}.$$

Therefore

$$y_1 + y_2 = 2\sqrt{3}.$$

But $$y_1 + y_2$$ is precisely $$Im(z_1 + z_2)$$, the imaginary part of the sum of the two complex numbers. So we conclude

$$Im(z_1 + z_2) = 2\sqrt{3}.$$

Hence, the correct answer is Option A.