The angle between the linee \(\bar{r}=(i+\hat{j}-\hat{k})+\lambda(3 i+\hat{j})\) and the plane \(2+3 \hat{k})=8\)

(D)
The angle between the line \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\) and the plane \(\overline{\mathrm{r}}-\overline{\mathrm{n}}=\mathrm{dr}\) is \(\sin \theta=\left|\frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}}{|\overline{\mathrm{b}}| \cdot|\overline{\mathrm{n}}|}\right|\)
Here \(\overline{\mathrm{b}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}\) and \(\overline{\mathrm{n}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\overline{\mathrm{b}} \cdot \overline{\mathrm{n}} =(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=3+2+0=5 \)
\( |\mathrm{~b}| =\sqrt{3^{2}+1^{2}}=\sqrt{10} \text { and }|\overline{\mathrm{n}}|=\sqrt{1^{2}+2^{2}+3^{2}}=\) \(\sqrt{14} \)
\( \sin \theta =\left|\frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}}{|\overline{\mathrm{b}}| \cdot|\overline{\mathrm{n}}|}\right|=\left|\frac{5}{\sqrt{10} \cdot \sqrt{14}}\right| \Rightarrow \sin \theta=\frac{\sqrt{5}}{2 \sqrt{7}} \Rightarrow\) \(\theta=\sin ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right)\)