= Solution
The <dot product> and <Euclidean norm> are $\mathbf x\cdot\mathbf y=\sum_i x_iy_i$ and $|\mathbf x|=\sqrt{\mathbf x\cdot\mathbf x}$. For every real $t$,
$$
0\leq|\mathbf x-t\mathbf y|^2=|\mathbf x|^2-2t\,\mathbf x\cdot\mathbf y+t^2|\mathbf y|^2.
$$
Its <quadratic discriminant> is nonpositive, proving the <Cauchy-Schwarz inequality> $|\mathbf x\cdot\mathbf y|\leq|\mathbf x||\mathbf y|$. In the inequality asked for, equality holds exactly when $\mathbf x$ is a positive scalar multiple of $\mathbf y$, and
$$
\boxed{\theta=\arccos\frac{\mathbf x\cdot\mathbf y}{|\mathbf x||\mathbf y|}.}
$$
Using the <Levi-Civita symbol> and its contraction identity,
$$
(\mathbf a\times\mathbf b)\cdot(\mathbf b\times\mathbf c)
=\varepsilon_{ijk}a_jb_k\varepsilon_{i\ell m}b_\ell c_m
=(\mathbf a\cdot\mathbf b)(\mathbf b\cdot\mathbf c)-|\mathbf b|^2(\mathbf a\cdot\mathbf c).
$$
Put $T=\mathbf a\cdot(\mathbf b\times\mathbf c)$, the <scalar triple product>. Cyclic symmetry gives $\mathbf m\cdot\mathbf a=\mathbf m\cdot\mathbf b=\mathbf m\cdot\mathbf c=T$, so each requested angle $\phi$ satisfies $\cos\phi=T/|\mathbf m|$. If every pairwise angle is $\theta$ and $t=\cos\theta$, the preceding identity gives
$$
|\mathbf m|^2=3(1-t^2)+6(t^2-t)=3(1-t)^2,
$$
and therefore \b[$|\mathbf m|=\sqrt3(1-\cos\theta)$]. This is $0$ when $\theta=0$; for a right-handed <orthonormal basis> and $\theta=\pi/2$, $\mathbf m=\mathbf a+\mathbf b+\mathbf c$ has norm $\sqrt3$.
The three <planes> have equations $\mathbf a\cdot\mathbf x=\mathbf a\cdot\mathbf p$, $\mathbf b\cdot\mathbf x=\mathbf b\cdot\mathbf q$, and $\mathbf c\cdot\mathbf x=\mathbf c\cdot\mathbf r$. They meet at one point exactly when $\Delta=\mathbf a\cdot(\mathbf b\times\mathbf c)\ne0$, meaning their normals are <linearly independent vectors>. The point is
$$
\boxed{\mathbf x=
\frac{(\mathbf a\cdot\mathbf p)(\mathbf b\times\mathbf c)+(\mathbf b\cdot\mathbf q)(\mathbf c\times\mathbf a)+(\mathbf c\cdot\mathbf r)(\mathbf a\times\mathbf b)}
{\mathbf a\cdot(\mathbf b\times\mathbf c)}.}
$$
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