The angle between the line \(\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}\) and plane \(x-2 y-\lambda z=3\) is \(\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\), then value of \(\lambda\) is

The acute angle \(\theta\) between line \(\bar{a}+\lambda \bar{b}\) and the plane \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\mathrm{p}\) is given by
\(\sin \theta=\left|\frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}}{|\overline{\mathrm{b}} \cdot \cdot \cdot \overrightarrow{\mathrm{n}}|}\right|\)
Here, \(\overline{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) and \(\overline{\mathrm{n}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}\)
Also, \(\theta=\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)=\sin ^{-1}\left(\frac{1}{3}\right)\)
\(\therefore \text { (i) } \Rightarrow \sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)=\) \(\left|\frac{2-2+2 \lambda}{\sqrt{4+1+4} \sqrt{1+4+\lambda^2}}\right| \)
\( \Rightarrow \frac{1}{3}=\left|\frac{2 \lambda}{3 \sqrt{5+\lambda^2}}\right| \)
\( \Rightarrow 5+\lambda^2=4 \lambda^2 \)
\( \Rightarrow \lambda^2=\frac{5}{3} \)
\( \Rightarrow \lambda=\sqrt{\frac{5}{3}}\)