If the locus of $$z \in \mathbb{C}$$, such that $$\text{Re}\left(\frac{z-1}{2z+i}\right) + \text{Re}\left(\frac{\bar{z}-1}{2\bar{z}-i}\right) = 2$$, is a circle of radius $$r$$ and center $$(a, b)$$ then $$\frac{15ab}{r^2}$$ is equal to :

Let $$z = x + iy$$, where $$x, y \in \mathbb{R}$$.

Define $$w = \dfrac{z-1}{2z+i}\,.$$
Its complex conjugate is $$\overline{w}= \dfrac{\bar z-1}{2\bar z-i}\,.$

The given condition is
$$$$\text{Re}$$\!$$\left$$(w$$\right$$)+$$\text{Re}$$\!$$\left$$($$\overline{w}$$$$\right$$)=2.$$

For any complex number $$w$$, $$$$\text{Re}$$(w)=$$\text{Re}$$($$\overline{w}$$)$$. Therefore
$$2\,$$\text{Re}$$(w)=2 \;\Longrightarrow\; $$\text{Re}$$(w)=1.$$

Hence we need the locus of $$z$$ satisfying
$$$$\text{Re}$$\!$$\left$$(\dfrac{z-1}{2z+i}$$\right$$)=1.$$

Write numerator and denominator in terms of $$x, y$$:
$$z-1=(x-1)+iy,$$
$$2z+i=2x+i(2y+1).$$

Multiply by the conjugate of the denominator to extract the real part:
$$\dfrac{z-1}{2z+i}= \dfrac{(x-1)+iy}{2x+i(2y+1)}\; $$\cdot$$\dfrac{2x-i(2y+1)}{2x-i(2y+1)}.$$

Denominator magnitude squared:
$$D=(2x)^2+(2y+1)^2=4x^2+(2y+1)^2.$$

Numerator product:
$$\;(x-1)+iy$$$$\cdot$$$$2x-i(2y+1)$$ gives

Real part $$R=2x(x-1)+y(2y+1)=2x^2-2x+2y^2+y.$$

Thus
$$$$\text{Re}$$\!$$\left$$(\dfrac{z-1}{2z+i}$$\right$$)=\dfrac{R}{D}=1 \;\Longrightarrow\; R=D.$$

Equate and rearrange:
$$2x^2-2x+2y^2+y =4x^2+4y^2+4y+1,$$
$$0=2x^2+2y^2+2x+3y+1.$$

Divide by 2 for simplicity:
$$x^2+y^2+x+\dfrac{3}{2}y+\dfrac{1}{2}=0.$$

Complete the squares:

$$x^2+x=$$\left$$(x+\dfrac{1}{2}$$\right$$)^2-\dfrac{1}{4},$$
$$y^2+\dfrac{3}{2}y=$$\left$$(y+\dfrac{3}{4}$$\right$$)^2-$$\left$$(\dfrac{3}{4}$$\right$$)^2.$$

Substitute:

$$$$\left$$(x+\dfrac{1}{2}$$\right$$)^2-\dfrac{1}{4} +$$\left$$(y+\dfrac{3}{4}$$\right$$)^2-\dfrac{9}{16} +\dfrac{1}{2}=0.$$

Combine constants:
$$-\dfrac{1}{4}-\dfrac{9}{16}+\dfrac{1}{2} =-\dfrac{5}{16}.$$

Move to the right side:

$$$$\left$$(x+\dfrac{1}{2}$$\right$$)^2+$$\left$$(y+\dfrac{3}{4}$$\right$$)^2 =\dfrac{5}{16}.$$

Hence the locus is a circle with
center $$\bigl(a,b\bigr)=$$\left$$(-\dfrac{1}{2},\,-\dfrac{3}{4}$$\right$$),$$
radius $$r=$$\sqrt{\dfrac{5}{16}$$}=\dfrac{$$\sqrt{5}$$}{4},$$
so $$r^{2}=\dfrac{5}{16}.$$

Now evaluate $$\dfrac{15ab}{r^{2}}.$$
$$ab=$$\left$$(-\dfrac{1}{2}$$\right$$)\!$$\left$$(-\dfrac{3}{4}$$\right$$)=\dfrac{3}{8},$$
$$\dfrac{15ab}{r^{2}}=\dfrac{15$$\cdot$$$$\frac{3}{8}$$}{$$\frac{5}{16}$$} =\dfrac{45}{8}$$\cdot$$\dfrac{16}{5}=18.$$

The required value is $$18$$, which corresponds to Option C.