Find the angle between the line $\overrightarrow{r} = \hat{i} + \hat{j} + 3 \hat{k} + \lambda \left(2 \hat{i} + \hat{j} - \hat{k}\right)$ and the plane $\overrightarrow{r} \cdot \left(\hat{i} + \hat{j}\right) = 1$
Equation of given line is: $\;$ $\overrightarrow{r} = \hat{i} + \hat{j} + 3 \hat{k} + \lambda \left(2 \hat{i} + \hat{j} - \hat{k}\right)$
It is of the form: $\;$ $\overrightarrow{r}= \overrightarrow{a} + \lambda \overrightarrow{b}$
Here, $\overrightarrow{b} = 2 \hat{i} + \hat{j} - \hat{k}$
Equation of the given plane is: $\;$ $\overrightarrow{r} \cdot \left(\hat{i} + \hat{j}\right) = 1$
Normal to the given plane is $\;$ $\overrightarrow{n} = \hat{i} + \hat{j}$
Let $\theta$ be the angle between the given line and the plane.
Then,
$\begin{aligned}
\sin \theta & = \dfrac{\overrightarrow{b} \cdot \overrightarrow{n}}{\left|\overrightarrow{b}\right| \; \left|\overrightarrow{n}\right|} \\\\
& = \dfrac{\left(2 \hat{i} + \hat{j} - \hat{k}\right) \cdot \left(\hat{i} + \hat{j}\right)}{\sqrt{\left(2\right)^2 + \left(1\right)^2 + \left(-1\right)^2} \; \sqrt{\left(1\right)^2 + \left(1\right)^2}} \\\\
& = \dfrac{3}{2 \sqrt{3}} = \dfrac{\sqrt{3}}{2}
\end{aligned}$
$\therefore$ $\;$ $\theta = \sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{3}$