Find the angle between the planes $\;$ $2x + y -z = 9$ $\;$ and $\;$ $x + 2y + z = 7$
The given planes are
$2x + y -z = 9$ $\;\;\; \cdots \; (1a)$
$x + 2y + z = 7$ $\;\;\; \cdots \; (2a)$
The normals to the given planes are
$\overrightarrow{n_1} = 2 \hat{i} + \hat{j} - \hat{k}$ $\;\;\; \cdots \; (1b)$
$\overrightarrow{n_2} = \hat{i} + 2 \hat{j} + \hat{k}$ $\;\;\; \cdots \; (2b)$
If $\;$ $\theta$ $\;$ is the angle between the planes, then
$\begin{aligned}
\cos \theta & = \dfrac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right| \left|\overrightarrow{n_2}\right|} \\\\
& = \dfrac{\left(2 \hat{i} + \hat{j} - \hat{k}\right) \cdot \left(\hat{i} + 2 \hat{j} + \hat{k}\right)}{\sqrt{\left(2\right)^2 + \left(1\right)^2 + \left(-1\right)^2} \; \sqrt{\left(1\right)^2 + \left(2\right)^2 + \left(1\right)^2}} \\\\
& = \dfrac{2 + 2 -1}{\sqrt{6} \; \sqrt{6}}
\end{aligned}$
$\therefore$ $\;$ $\theta = \cos^{-1} \left(\dfrac{3}{6}\right) = \dfrac{\pi}{3}$