The angles of a triangle are in AP and the number of degree in the least is to the number of radians in the greatest is $60:\pi .$ If the difference between least and greatest angles is ${\left(\frac{\lambda \pi }{6}\right)}^C,$ then the value of $\lambda$ is

Le a-d,a,a+d are angles in degree of a triangle in AP, then
$a-d+a+a+d={180}^0$ 
$3a={180}^0\Rightarrow a={60}^0$ 
Greatest angle in radian $=(a+d)\times \frac{\pi }{180}rad$ 
According to the question, $\frac{a-d}{\frac{\pi }{180}(a+d)}=\frac{60}{\pi }$
or $3(a-d)=a+d$  
$\Rightarrow d=\frac{1}{2}a=\frac{1}{2}\times {60}^0={30}^0$ 
$\therefore$ Angles are ${60}^0-{30}^0,{60}^0,{60}^0+{30}^0={30}^0,{60}^0,{90}^0\Rightarrow \frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{2}$  
According to the question, $\frac{\pi }{2}-\frac{\pi }{6}=\frac{\lambda \pi }{6}\Rightarrow \frac{2\pi }{6}=\frac{\lambda \pi }{6}\Rightarrow \lambda =2$