If $\theta $ is the angle between the line $\vec{r}= (\hat{i} + 2\hat{j} - \hat{k}) + λ(\hat{i} -\hat{j} + \hat{k})$ and the plane $\vec{r}.(2\hat{i} -\hat{j} + \hat{k})=4$, then cos $\theta $ = 

none of these

We know that the angle $\theta $ between the line $\vec{r}=\vec{a}+λ\vec{b}$ and the plane $\vec{r}.\vec{n} = d $ is given by

$sin \theta = \frac{\vec{b}.\vec{n}}{|\vec{b}||\vec{n}|}$

Here, $\vec{b}=\hat{i}-\hat{j} + \hat{k}$ and $\vec{n} = 2\hat{i}-\hat{j} + \hat{k}.$

$∴sin \theta = \frac{(\hat{i}-\hat{j} + \hat{k}).(2\hat{i}-\hat{j} + \hat{k})}{\sqrt{1^2+(-1)^2+1^2}\sqrt{2^2+(-1)^2+1^2}}$

$⇒ sin \theta = \frac{2+1+1}{\sqrt{3}\sqrt{6}}=\frac{4}{3\sqrt{2}}=\frac{2\sqrt{2}}{3}⇒ cos \theta = \frac{1}{3}$