• # question_answer If the angle 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$\theta$such that$\sin \theta =\frac{1}{3},$then the value of$\lambda$is, A)  $\frac{5}{3}$ B)  $-\frac{3}{5}$ C)  $\frac{3}{4}$ D)  $-\frac{4}{3}$

For the line$\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ ${{a}_{1}}=1,{{b}_{1}}=2,{{c}_{1}}=2$ and for the plane$2x-y+\sqrt{\lambda }z+4=0$ ${{a}_{2}}=2,{{b}_{2}}=-1,{{c}_{2}}=\sqrt{\lambda }$ $\therefore$$\sin \theta =\frac{1}{3}=\left| \frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \right|$ $\Rightarrow$ $\frac{1}{3}=\left| \frac{2-2+2\sqrt{\lambda }}{\sqrt{1+4+4}\sqrt{4+1+\lambda }} \right|$ $\Rightarrow$ $\frac{1}{3}=\frac{2\sqrt{\lambda }}{\sqrt{9}\sqrt{5+\lambda }}$ $\Rightarrow$ $\sqrt{5+\lambda }=2\sqrt{\lambda }$ $\Rightarrow$ $5+\lambda =4\lambda$ $\Rightarrow$ $3\lambda =5$ $\Rightarrow$ $\lambda =\frac{5}{3}$