= Solution
The <vector> $\mathbf a$ is <orthogonal> to both $\mathbf b$ and $\mathbf c$, so $\mathbf a\cdot(\mathbf b-\mathbf c)=0$. Also $\mathbf d\cdot\mathbf b=\mathbf d\cdot\mathbf c=1/S$, hence $\mathbf d\cdot(\mathbf b-\mathbf c)=0$. Subtraction gives $(\mathbf a-\mathbf d)\cdot(\mathbf b-\mathbf c)=0$. Both vectors are nonzero, so
$$
\boxed{\text{the angle is }\pi/2.}
$$
This is the altitude property of the <orthogonal projection onto an intercept plane>: the projection point is the triangle's <orthocenter>.
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