The angle between the line
\(\overline{\mathrm{r}}=(\hat{\imath}+2 \hat{\mathrm{\jmath}}-\widehat{\mathrm{k}})+\lambda(\hat{\imath}-\hat{\mathrm{\jmath}}+\widehat{\mathrm{k}})\) and the plane
\(\overline{\mathrm{r}} \cdot(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})=4 \mathrm{is}\)

The angle \(\theta\) between the line \(\bar{r}=\bar{a}+\lambda b\) and the plane \(\bar{r} \cdot \bar{n}=p\) is given by
\(\sin \theta=\frac{\bar{b} \cdot \bar{n}}{|\bar{b}| \cdot|\bar{n}|}\)
Here \(\overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\overline{\mathrm{n}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\therefore \overline{\mathrm{b}} \cdot \overline{\mathrm{n}}=1(2)+(-1)(-1)+1(1)=4\)
\(|\bar{b}|=\sqrt{1+1+1}=\sqrt{3}\) and \(|\bar{n}|=\sqrt{4+1+1}=\sqrt{6}=\sqrt{3} \times \sqrt{2}\)
\(\therefore \sin \theta=\frac{4}{\sqrt{3} \times \sqrt{3} \times \sqrt{2}}=\frac{2 \sqrt{2}}{3} \Rightarrow \theta=\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\)