Find the angle between the line $\vec{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$ and the plane $\vec{r}\cdot (2\hat{i}-\hat{j}+\hat{k})=6$.

Let $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be the direction vector of the line.Let $\vec{n}=2\hat{i}-\hat{j}+\hat{k}$ be the normal vector to the plane.The angle $\theta$ between a line and a plane is given by:
$$\sin \theta =\frac{|\vec{b}\cdot \vec{n}|}{|\vec{b}||\vec{n}|}$$
Calculate dot product: $\vec{b}\cdot \vec{n}=(1)(2)+(-1)(-1)+(1)(1)=2+1+1=4$.Calculate magnitudes:$|\vec{b}|=\sqrt{1^2+(-1{)}^2+1^2}=\sqrt{3}$$|\vec{n}|=\sqrt{2^2+(-1{)}^2+1^2}=\sqrt{6}$ Calculate $\sin \theta$:
$$\sin \theta =\frac{4}{\sqrt{3}\sqrt{6}}=\frac{4}{\sqrt{18}}=\frac{4}{3\sqrt{2}}=\frac{2\sqrt{2}}{3}$$
$$\theta ={\sin }^{-1}(\frac{2\sqrt{2}}{3})$$