= Solution
Let $\mathbf p=(2,1,1)$, $\mathbf q=(1,2,2)$ and $\mathbf r=(0,0,3)$. <Cross products> give plane normals $\mathbf p\times\mathbf r=(3,-6,0)$ and $\mathbf q\times\mathbf r=(6,-3,0)$. Their dot product is $36$ and each has norm $3\sqrt5$. Thus the acute <angle between two planes> is
$$
\boxed{\theta=\arccos\frac{36}{45}=\arccos\frac45\simeq36.87^\circ.}
$$
This is also the interior dihedral angle here: perpendicular to the common edge $OR$, the interior wedge lies between the projected directions $(2,1)$ and $(1,2)$. The two outward face normals instead make the supplementary angle.
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