= Solution
For a differentiable curve $R(t)$ in the <SO(3) group> with $R(0)=I$, differentiating $R(t)^TR(t)=I$ gives $X^T+X=0$, where $X=R'(0)$. Conversely, if $X$ is a real <skew-symmetric matrix>, $e^{tX}$ is an <orthogonal matrix> and has <determinant> $e^{t\operatorname{tr}X}=1$. Hence
$$
\boxed{\mathfrak{so}(3)=\{X\in M_3(\mathbb R):X^T=-X\},\qquad\dim\mathfrak{so}(3)=3.}
$$
The displayed generators of the <SO(3) Lie algebra> are
$$
T_1=\begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix},\quad T_2=\begin{pmatrix}0&0&1\\0&0&0\\-1&0&0\end{pmatrix},\quad T_3=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}.
$$
They are <linearly independent> and span the three independent entries of a real <skew-symmetric matrix>. Equivalently, $T_a\mathbf v=\mathbf e_a\times\mathbf v$. The vector triple-product identity then gives
$$
[T_a,T_b]\mathbf v=\mathbf e_a\times(\mathbf e_b\times\mathbf v)-\mathbf e_b\times(\mathbf e_a\times\mathbf v)=(\mathbf e_a\times\mathbf e_b)\times\mathbf v.
$$
Thus the <structure constants of a Lie algebra> in this basis are
$$
\boxed{[T_a,T_b]=\epsilon_{abc}T_c,\qquad f_{ab}{}^c=\epsilon_{abc}.}
$$
Use a real, antisymmetric convention throughout the <SO(3) vector Higgs model>. With <gauge coupling> $g$, the <gauge covariant derivative> and <gauge field strength> are
$$
\boxed{D_\mu\Phi=\partial_\mu\Phi+gA_\mu\Phi,\qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+g[A_\mu,A_\nu],\qquad[D_\mu,D_\nu]=gF_{\mu\nu}.}
$$
There is no factor of $i$ in these formulas because $T_a$ are real antisymmetric generators. For a local <non-Abelian gauge transformation> $\Phi'=R\Phi$, with $R(x)$ in the <SO(3) group>, the compatible transformation of the <gauge potential> is
$$
A'_\mu=RA_\mu R^{-1}-g^{-1}(\partial_\mu R)R^{-1}.
$$
It gives $D'_\mu\Phi'=R D_\mu\Phi$ and $F'_{\mu\nu}=RF_{\mu\nu}R^{-1}$. These covariance laws fix the relative signs in the <gauge covariant derivative> and <gauge field strength>.
An <SO(3) invariant quartic scalar potential> is
$$
\boxed{V(\Phi)=\frac\lambda4(\Phi^T\Phi-v^2)^2,\qquad\lambda>0,\quad v>0.}
$$
Since $(R\Phi)^T(R\Phi)=\Phi^T\Phi$, this <scalar potential> is <gauge-invariant>. Its minimum is the sphere $\Phi^T\Phi=v^2$, rather than a preferred direction in the internal vector space.
Writing $\mathbf A_\mu=(A_{\mu1},A_{\mu2},A_{\mu3})$, the <cross product> form of the <gauge covariant derivative> is $D_\mu\Phi=\partial_\mu\Phi+g\mathbf A_\mu\times\Phi$. In components this is
$$
\boxed{\begin{aligned}(D_\mu\Phi)_1&=\partial_\mu\Phi_1+g(A_{\mu2}\Phi_3-A_{\mu3}\Phi_2),\\(D_\mu\Phi)_2&=\partial_\mu\Phi_2+g(A_{\mu3}\Phi_1-A_{\mu1}\Phi_3),\\(D_\mu\Phi)_3&=\partial_\mu\Phi_3+g(A_{\mu1}\Phi_2-A_{\mu2}\Phi_1).\end{aligned}}
$$
Correspondingly, the <gauge field strength> has components
$$
F_{\mu\nu a}=\partial_\mu A_{\nu a}-\partial_\nu A_{\mu a}+g\epsilon_{abc}A_{\mu b}A_{\nu c}.
$$
A <covariantly constant vector Higgs field> has constant norm, because
$$
\partial_\mu(\Phi^T\Phi)=2\Phi^T\partial_\mu\Phi=-2g\Phi^TA_\mu\Phi=0.
$$
The last equality uses antisymmetry of the <gauge potential>. On a connected region write this norm as $\phi\geq0$. If $\phi=0$, the <vector Higgs field> is identically zero. If $\phi>0$, the <SO(3) group> acts transitively on <unit vectors>, so choose a smooth local <non-Abelian gauge transformation> rotating $\Phi/\phi$ to $\mathbf e_3$. In this <unitary gauge>, $\Phi=(0,0,\phi)^T$ with $\phi$ constant. This is a local construction, and is global on a trivial contractible region; a nontrivial bundle can require more than one gauge patch.
For nonzero $\phi$, substituting the constant aligned <vector Higgs field> into $D_\mu\Phi=0$ gives $g\phi(A_{\mu2},-A_{\mu1},0)^T=0$. For nonzero <gauge coupling>, \b[the most general compatible <gauge potential> and <gauge field strength> are]
$$
\boxed{A_\mu=a_\mu T_3,\qquad F_{\mu\nu}=(\partial_\mu a_\nu-\partial_\nu a_\mu)T_3,}
$$
where $a_\mu$ is an arbitrary real one-form. Thus the surviving <gauge group> is the $SO(2)$ <stabilizer subgroup> of $\mathbf e_3$. A <covariantly constant vector Higgs field> does not force the <gauge field strength> to vanish: it only requires the curvature to annihilate that field.
For the <Higgs mechanism>, choose a vacuum of the <SO(3) invariant quartic scalar potential>, so the constant norm is $v$. The canonically normalized <kinetic term> then contains
$$
\frac12(D_\mu\Phi)^T(D^\mu\Phi)\supset\frac12g^2v^2\left(A_{\mu1}A_1^\mu+A_{\mu2}A_2^\mu\right).
$$
The two broken-direction <gauge bosons> have mass $gv$ (or $|g|v$ if the sign of $g$ is unrestricted), while the $T_3$ <gauge boson> remains massless. The two angular <Goldstone bosons> supply the massive vectors' <longitudinal gauge-boson polarizations>. The radial fluctuation remains a physical <scalar field>, with $m_h^2=2\lambda v^2$ for this normalization. The condition $D_\mu\Phi=0$ selects backgrounds without broken-direction <gauge potentials>; fluctuations about such a vacuum exhibit the two massive modes. Constancy of the norm alone does not require it to minimize the <scalar potential>.
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