If the angle between the line $\mathrm{x}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-3}{\mathrm{\lambda }}$ and the plane $\mathrm{x}+2\mathrm{y}+3\mathrm{z}=4$ is ${\cos }^{-1}\left(\sqrt{5/14}\right)\text{ then }\mathrm{\lambda }=$

Let ${\cos }^{-1}\left(\sqrt{\frac{5}{14}}\right)=\mathrm{\theta }\Rightarrow \cos \mathrm{\theta }=\frac{\sqrt{5}}{\sqrt{14}}$$\Rightarrow \sin \mathrm{\theta }=\frac{3}{\sqrt{14}}$

D.r’s of given line are $\left({\mathrm{a}}_1,{\mathrm{b}}_1,{\mathrm{c}}_1\right)=(1,2,\mathrm{\lambda })$ 

$d.r\text{'}s of normal to the given plane are \left({\mathrm{a}}_2,{\mathrm{b}}_2,{\mathrm{c}}_2\right)=(1,2,3)$
$Now \sin \mathrm{\theta }=\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2}{\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2+{\mathrm{c}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2+{\mathrm{c}}_2^2}}$
$$\begin{gathered} \frac{3}{\sqrt{14}}=\frac{1+4+3\mathrm{\lambda }}{\sqrt{1+4+{\mathrm{\lambda }}^2}\sqrt{1+4+9}} \\ \Rightarrow 9(5+{\lambda }^2)={(5+3\lambda )}^2 \Rightarrow 45+9{\lambda }^2=25+9{\lambda }^2+30\lambda \\ \Rightarrow 30\lambda =20 \\ \Rightarrow \mathrm{\lambda }=\frac{2}{3} \end{gathered}$$