If $\mathrm{z}=\frac{(1+\mathrm{i})(1+2\mathrm{i})(1+3\mathrm{i})}{(1-\mathrm{i})(2-\mathrm{i})(3-\mathrm{i})}$ then, argument of z is ____

$\begin{array}{l}\arg \mathrm{z}=\arg (1+\mathrm{i})+\arg (1+2\mathrm{i})+\arg (1+3\mathrm{i})\\ -\arg (1-\mathrm{i})-\arg (2-\mathrm{i})-\arg (3-\mathrm{i})\\ ={\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(3\right)\\ -{\tan }^{-1}(-1)-{\tan }^{-1}\left(\frac{-1}{2}\right)-{\tan }^{-1}\left(\frac{-1}{3}\right)\\ =\mathrm{\pi }+\frac{\mathrm{\pi }}{2}=\frac{3\mathrm{\pi }}{2}\end{array}$