The value of $\lambda$ for which the loci $\arg z=\frac{\pi }{6}$ and $|z-2\sqrt{3}i|=\lambda$ on the argand plane touch each other is 

$z=x+iy$

$\arg z={\tan }^{-1}\left(\frac{y}{x}\right)$

$\arg \left(z\right)=\frac{\pi }{6}$

$\therefore {\tan }^{-1}\frac{y}{x}=\frac{\pi }{6}$

$\frac{y}{x}=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$

$\Rightarrow x=\sqrt{3}y$

Also given, 

$|z-2\sqrt{3}i|=\lambda$

$|x+iy-2\sqrt{3}i|=\lambda$

$x^2+{\left(y-2\sqrt{3}\right)}^2={\lambda }^2$

This is the equation of circle whose centre $x-\sqrt{3}y=0$ is $(0,2\sqrt{3})$ and $r=\lambda$. Now, the line touches the circle

$\therefore \lambda =\left|\frac{0-\sqrt{3}\left(2\sqrt{3}\right)}{\sqrt{1+3}}\right|$

$\lambda =\left|\frac{6}{2}\right|=3$

$\therefore \lambda =3$