Let $$z = 1 + i$$ and $$z_1 = \frac{1 + i\bar{z}}{\bar{z}(1-z) + \frac{1}{z}}$$. Then $$\frac{12}{\pi} \arg z_1$$ is equal to

We are given $$z = 1 + i$$ and asked to find $$\frac{12}{\pi}\arg(z_1)$$ where $$z_1 = \frac{1 + i\bar{z}}{\bar{z}(1-z) + \frac{1}{z}}$$. To begin, we note that $$\bar{z} = 1 - i$$, so $$1 + i\bar{z} = 1 + i(1 - i) = 1 + i - i^2 = 1 + i + 1 = 2 + i$$.

Next, we calculate the denominator. First, $$\bar{z}(1 - z) = (1 - i)(1 - 1 - i) = (1 - i)(-i) = -i + i^2 = -1 - i$$, and $$\frac{1}{z} = \frac{1}{1 + i} = \frac{1 - i}{2}$$. Adding these gives $$-1 - i + \frac{1 - i}{2} = \frac{-2 - 2i + 1 - i}{2} = \frac{-1 - 3i}{2}$$.

Therefore, $$z_1 = \frac{2 + i}{\frac{-1 - 3i}{2}} = \frac{2(2 + i)}{-1 - 3i} = \frac{4 + 2i}{-1 - 3i}$$. Multiplying numerator and denominator by the conjugate of the denominator yields $$z_1 = \frac{(4 + 2i)(-1 + 3i)}{(-1 - 3i)(-1 + 3i)} = \frac{-4 + 12i - 2i + 6i^2}{1 + 9} = \frac{-4 + 10i - 6}{10} = \frac{-10 + 10i}{10} = -1 + i\,. $$

Since $$\arg(-1 + i) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$, it follows that $$\frac{12}{\pi} \times \frac{3\pi}{4} = \frac{12 \times 3}{4} = 9$$. Thus, the answer is $$\boxed{9}$$.