Given planes are:
$\bar{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=3$ and $\bar{r} \cdot(\hat{i}+2 \hat{j}+\widehat{k})=1$
The angle between two planes with direction ratios,
$\left(a_1, b_1, c_1\right)$ and $\left(a_2, b_2, c_2\right)$ is
$\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$
$\cos \theta=\frac{2 \times 1+1 \times 2-1 \times 1}{\sqrt{2^2+1^2+(-1)^2} \sqrt{1^2+2^2+1^2}}$
$\cos \theta=\frac{3}{\sqrt{6} \cdot \sqrt{6}}$
$\cos \theta=\frac{3}{6}$
$\cos \theta=\frac{1}{2}$
$\cos \theta=\cos \left(\frac{\pi}{3}\right)$
$\theta=\frac{\pi}{3}$
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