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

Direction ratios of line are (a₁, b₁,c₁) = (1,2,2)

And direction ratios of a plane are $\left({\mathrm{a}}_2,{\mathrm{b}}_2,{\mathrm{c}}_2\right)=(2,-1,\sqrt{\mathrm{\lambda }})$

$\begin{array}{l}\therefore \sin \mathrm{\theta }=\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2}{\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2+{\mathrm{c}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2+{\mathrm{c}}_2^2}}\\ =\frac{1\times 2+2(-1)+2\times \sqrt{\mathrm{\lambda }}}{\sqrt{1^2+2^2+2^2}\sqrt{2^2+1^2+(\sqrt{\mathrm{\lambda }}{)}^2}}\\ \Rightarrow \frac{1}{3}=\frac{2\sqrt{\mathrm{\lambda }}}{3\sqrt{5+\mathrm{\lambda }}}\Rightarrow 5+\mathrm{\lambda }=4\mathrm{\lambda }\Rightarrow \mathrm{\lambda }=\frac{5}{3}\end{array}$