If $$z$$ and $$\omega$$ are two complex numbers such that $$|z\omega| = 1$$ and $$\arg(z) - \arg(\omega) = \frac{3\pi}{2}$$, then $$\arg\left(\frac{1 - 2\bar{z}\omega}{1 + 3\bar{z}\omega}\right)$$ is:
(Here $$\arg(z)$$ denotes the principal argument of complex number $$z$$)

We are given $$|z\omega| = 1$$ and $$\arg(z) - \arg(\omega) = \frac{3\pi}{2}$$.

Let $$\arg(z) = \alpha$$ and $$\arg(\omega) = \beta$$. Since $$\bar{z}$$ has argument $$-\alpha$$, we get $$\arg(\bar{z}\omega) = -\alpha + \beta = -(\alpha - \beta) = -\frac{3\pi}{2}$$. Adding $$2\pi$$ to bring this into the principal range $$(-\pi, \pi]$$ gives $$\arg(\bar{z}\omega) = \frac{\pi}{2}$$. Also, $$|\bar{z}\omega| = |z\omega| = 1$$. Therefore $$\bar{z}\omega = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i$$.

Substituting $$\bar{z}\omega = i$$: $$\frac{1 - 2\bar{z}\omega}{1 + 3\bar{z}\omega} = \frac{1 - 2i}{1 + 3i}.$$

Multiplying numerator and denominator by the conjugate of the denominator: $$\frac{1 - 2i}{1 + 3i} \cdot \frac{1 - 3i}{1 - 3i} = \frac{(1)(1) + (1)(-3i) + (-2i)(1) + (-2i)(-3i)}{1^2 + 3^2} = \frac{1 - 3i - 2i + 6i^2}{10} = \frac{1 - 5i - 6}{10} = \frac{-5 - 5i}{10} = \frac{-1 - i}{2}.$$

The complex number $$w = \frac{-1-i}{2}$$ lies in the third quadrant (both real and imaginary parts are negative). Its reference angle is $$\arctan\!\left(\frac{1/2}{1/2}\right) = \frac{\pi}{4}$$. The principal argument (in $$(-\pi, \pi]$$) is therefore $$-\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$$.

Therefore $$\arg\!\left(\frac{1 - 2\bar{z}\omega}{1 + 3\bar{z}\omega}\right) = -\dfrac{3\pi}{4}$$.