= Solution
The <normal vectors> are $\mathbf n_A=(1,2,-4)$ and $\mathbf n_B=(2,-1,3)$. Their <dot product> is $-12$, and their <norms> are $\sqrt{21}$ and $\sqrt{14}$. The usual <angle between two planes> is the smaller, unoriented angle, so
$$
\boxed{\theta=\arccos\frac{12}{\sqrt{21}\sqrt{14}}
=\arccos\frac{2\sqrt6}{7}\approx45.6^\circ.}
$$
For the particular signed <normal vectors> printed above, the angle is the supplement $\pi-\theta\approx134.4^\circ$. Reversing either <normal vector> exchanges these two angles without changing either <plane>.
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