If angle $\theta$ between the line  $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2\mathrm{x}-\mathrm{y}+\sqrt{\mathrm{\lambda }}\mathrm{z}+4=0$ is such that  $\sin \theta =\frac{1}{3},$the value of $3\lambda$ is

The line is $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane is  $2\mathrm{x}-\mathrm{y}+\sqrt{\mathrm{\lambda }}\mathrm{z}+4=0$

If $\theta$ is the angle between the line and the plane, then 90^o - $\theta$ is the angle between the line and normal to the plane. 

Thus, 

$\begin{array}{l}\cos \left({90}^{\circ }-\mathrm{\theta }\right)=\frac{\left(1\right)\left(2\right)+\left(2\right)(-1)+\left(2\right)\left(\sqrt{\mathrm{\lambda }}\right)}{\sqrt{1+4+4}\sqrt{4+1+\mathrm{\lambda }}}\\ \mathrm{or} \sin \mathrm{\theta }=\frac{2-2+2\sqrt{\mathrm{\lambda }}}{3\sqrt{5+\mathrm{\lambda }}}\text{ or }\frac{1}{3}=\frac{2\sqrt{\mathrm{\lambda }}}{3\sqrt{5+\mathrm{\lambda }}}\\ \mathrm{or} \sqrt{5+\mathrm{\lambda }}=2\sqrt{\mathrm{\lambda }}\\ \mathrm{or} 5+\mathrm{\lambda }=4\mathrm{\lambda }\\ \mathrm{or} 3\mathrm{\lambda }=5\\ \mathrm{or} \mathrm{\lambda }=\frac{5}{3}\end{array}$