If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$

and the plane $2x-y+\sqrt{\lambda }z+4=0$is such that $\sin \theta =\frac{1}{3},$ than $\lambda$ is

The $\text{ d.r's }$of the line are 1,2,2.

The $\text{ d.r's }$ of normal to the plane are $2,-1,\sqrt{\lambda }$

$\frac{1}{3}=\sin \theta =\frac{1\times 2+2\times (-1)+2\sqrt{\lambda }}{\sqrt{1+4+4}\sqrt{5+\lambda }}$

$\Rightarrow 6\sqrt{\lambda }=3\sqrt{5+\lambda }\Rightarrow 36\lambda =9(5+\lambda )$

$\Rightarrow 3\lambda =5\Rightarrow \lambda =\frac{5}{3}$.