= Solution
There is a sign inconsistency in the printed model. For the ordinary <cross product> let $J(a)z=a\times z$. The vector triple-product identity gives $[J(a),J(b)]=J(a\times b)$. Consequently the printed $D_\mu=\partial_\mu+eJ(A_\mu)$ has <gauge curvature>
$$
[D_\mu,D_\nu]\Phi=eF^+_{\mu\nu}\times\Phi,\qquad
F^+_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+eA_\mu\times A_\nu.
$$
Its <gauge curvature> cannot instead contain the printed negative quadratic term. For a noncommuting <pure gauge potential> for this <gauge covariant derivative>, $F^+=0$ whereas $F^-=-2eA_\mu\times A_\nu\ne0$. Thus the printed gauge <kinetic term> assigns a nonzero value to a connection obtained by applying the <gauge-field transformation law> to the zero connection. Literally the printed action has global rotations but not the requested local triplet <gauge symmetry>. The <cross-product convention for an adjoint covariant derivative> makes the minimal repair explicit: change the <gauge field strength> quadratic term to a plus sign, retaining the printed <gauge covariant derivative>. Changing the sign in the <gauge covariant derivative> instead would be another consistent convention. The remaining calculation uses the first repair.
The corrected theory has local <SO(3)> <gauge symmetry>, equivalently the adjoint-field realization of the <SU(2)> <Lie algebra>; its <adjoint scalar fields> and <gauge fields> are insensitive to the <center of a group> of <SU(2)>. Explicitly, $\Phi\mapsto R(x)\Phi$ and
$$
eJ(A_\mu)\mapsto R\,eJ(A_\mu)R^{-1}-(\partial_\mu R)R^{-1}.
$$
Both $D_\mu\Phi$ and $F_{\mu\nu}$ transform as triplets, so their <dot products> and the <scalar potential> are invariant. Assume $\lambda>0$, $e\ne0$ and take $v\ge0$ without loss of generality. For $v>0$ the minima have $\Phi^2=v^2$. Choose the <scalar-field vacuum> $\Phi_0=v(0,0,1)$, whose <stabilizer> consists of rotations around the third axis: \b[the residual <gauge symmetry> is SO(2), locally <U(1)>.]
In <unitary gauge> write $\Phi=(0,0,v+h)$, $A_\mu=A^3_\mu$ and $W^\pm_\mu=(A^1_\mu\mp iA^2_\mu)/\sqrt2$. These are the neutral massless <gauge field>, two charged massive <gauge fields> and a real radial <scalar field>. Define
$$
f_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu,\qquad
C_{\mu\nu}=W^+_\mu W^-_\nu-W^+_\nu W^-_\mu,
$$
$$
G^\pm_{\mu\nu}=(\partial_\mu\mp ieA_\mu)W^\pm_\nu-(\partial_\nu\mp ieA_\nu)W^\pm_\mu.
$$
Substitution gives $F^3_{\mu\nu}=f_{\mu\nu}-ieC_{\mu\nu}$ and $(F^1_{\mu\nu}\mp iF^2_{\mu\nu})/\sqrt2=G^\pm_{\mu\nu}$. The entire physical-field <Lagrangian density> is therefore
$$
\mathcal L=-\frac14(f_{\mu\nu}-ieC_{\mu\nu})(f^{\mu\nu}-ieC^{\mu\nu})
-\frac12G^+_{\mu\nu}G^{-\mu\nu}
+\frac12\partial_\mu h\partial^\mu h+e^2(v+h)^2W^+_\mu W^{-\mu}
-\frac\lambda8[(v+h)^2-v^2]^2.
$$
In particular $D_\mu\Phi=(e(v+h)A^2_\mu,-e(v+h)A^1_\mu,\partial_\mu h)$; this directly verifies the <scalar field> kinetic and vector <mass terms>. Expanding the <scalar potential> gives $\lambda v^2h^2/2+\lambda vh^3/2+\lambda h^4/8$. Comparing the quadratic terms with standard <real scalar field> and charged-vector <mass terms> yields the <adjoint triplet Higgs spectrum>,
$$
\boxed{m_{W^\pm}^2=e^2v^2,\qquad m_A=0,\qquad m_h^2=\lambda v^2}.
$$
This also rewrites the purely vector interactions, which are already contained in the displayed <gauge field strengths> rather than omitted from the theory.
The original symmetry remains unbroken \b[at $v=0$]. The unique minimum is then $\Phi=0$, with three massless <gauge fields> and three massless <real scalar field> components; the quartic <scalar field> interaction remains. The physical <degrees of freedom> number $3\times2+3=9$. At $v>0$ the two broken generators supply two <Goldstone bosons>, eaten to give <longitudinal gauge-boson polarizations> of the charged vectors. The physical count is $2\times3+2+1=9$: two massive vectors, one massless vector and one radial <scalar field>. The scalar components eaten in <unitary gauge> are not additional physical particles.
For the broken phase all interactions containing the physical <scalar field> are
$$
\mathcal L_{\rm scalar,int}=-\frac{\lambda v}{2}h^3-\frac\lambda8h^4
+2e^2vhW^+_\mu W^{-\mu}+e^2h^2W^+_\mu W^{-\mu}.
$$
Multiplying interaction coefficients by $i$ and differentiating with respect to the external fields includes the identical-field <factorials>. The <scalar vertices of an adjoint triplet Higgs model> are
$$
\boxed{hhh:\ -3i\lambda v,\quad hhhh:\ -3i\lambda,\quad
hW^+_\mu W^-_\nu:\ 2ie^2v g_{\mu\nu},\quad
hhW^+_\mu W^-_\nu:\ 2ie^2g_{\mu\nu}}.
$$
For example, the $h^3$ rule contains $3!$, whereas the $hhW^+W^-$ rule contains only $2!$ from its two identical <scalar fields>. There are no $hAA$ or $hhAA$ vertices: the radial <scalar field> is neutral under the residual <U(1)>. There is no derivative scalar-vector vertex in this physical spectrum in <unitary gauge>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-53-scalar-vertices.png]
{title=All physical-scalar interaction vertices in the broken triplet Higgs model, with the unbroken-phase scalar vertices shown below; dashed lines are scalars and wavy lines are vectors}
For completeness, if the unbroken phase is chosen, its physical <scalar fields> are the three components $\phi_a$, not a single radial $h$. Expanding its <gauge-covariant kinetic term> gives
$$
\mathcal L_{A\phi\phi}=e\epsilon^{iaj}(\partial_\mu\phi_i)A^{a\mu}\phi_j,\qquad
\mathcal L_{AA\phi\phi}=\frac{e^2}{2}[A_\mu^aA^{a\mu}\phi_b\phi_b-A_\mu^a\phi_aA^{b\mu}\phi_b].
$$
With all momenta incoming and <Fourier transform> convention $e^{-ipx}$, the three unbroken-phase rules are
$$
A^a_\mu\phi_b(p)\phi_c(q):\quad e\epsilon^{abc}(q-p)_\mu,
$$
$$
A^a_\mu A^b_\nu\phi_c\phi_d:\quad
ie^2(2\delta^{ab}\delta^{cd}-\delta^{ac}\delta^{bd}-\delta^{ad}\delta^{bc})g_{\mu\nu},
$$
$$
\phi_a\phi_b\phi_c\phi_d:\quad
-i\lambda(\delta^{ab}\delta^{cd}+\delta^{ac}\delta^{bd}+\delta^{ad}\delta^{bc}).
$$
The first rule follows by differentiating once on each possible scalar: for $a=3,b=1,c=2$, the interaction is $eA^3_\mu(\phi_1\partial^\mu\phi_2-\phi_2\partial^\mu\phi_1)$, which fixes the momentum sign. The remaining rules follow from the quadratic gauge interaction and $-\lambda(\phi_a\phi_a)^2/8$. There are no unbroken-phase cubic <scalar field> <Feynman vertices>.
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