1. Algebraic Approach:
- Let $z = a + bi$, where $a$ and $b$ are real numbers.
- Substitute $z$ and its conjugate $\overline{z} = a - bi$ into the given equation:
$$(1-2i)z + (-2i-1)\overline{z} = 6i$$
$$(1-2i)(a+bi) + (-2i-1)(a-bi) = 6i$$
- Expand and simplify:
$$(a - 2ai + b(-2i)^2) + (-a - 2ai - b(-2i)^2) = 6i$$
$$(a - 2ai - 2b + 4bi) + (-a - 2ai + 2b - 4bi) = 6i$$
$$-4ai + 4bi = 6i$$
$$(-4a + 4b)i = 6i$$
- Equate the imaginary parts:
$$-4a + 4b = 6$$
$$-4a + 4b = 6 \implies -4a + 4b = 6 \implies a - b = -\frac{3}{2}$$
- The real parts cancel out, leaving:
$$a^2 + b^2 + \frac{2}{3}a + \frac{1}{3}b = 0$$
- Complete the square:
$$a^2 + \frac{2}{3}a + b^2 + \frac{1}{3}b = 0$$
$$\left(a + \frac{1}{3}\right)^2 + \left(b + \frac{1}{6}\right)^2 = \frac{5}{36}$$
- This represents a circle with radius $\sqrt{\frac{5}{36}} = \frac{\sqrt{5}}{6}$.

2. Geometric Approach:
- Let $w = (1-2i)z$, then the equation becomes:
$$w - \overline{w} = 6i$$
- This represents a line in the complex plane where the imaginary part of $w$ is 3:
$$\text{Im}(w) = 3$$
- Inverting $w \mapsto \frac{1}{w}$ yields a circle centered at $\left(0, \frac{1}{6}\right)$ with radius $\frac{1}{6}$.
- The area of this circle is:
$$\pi \left(\frac{1}{6}\right)^2 = \frac{\pi}{36}$$
- Scaling by $\frac{|w|^2}{|z|^2} = 5$ gives:
$$5 \times \frac{\pi}{36} = \frac{5\pi}{36}$$

The final answer is $\boxed{\frac{5\pi}{36}}$.