= Solution
Assume $\lambda>0$ and $v>0$. The minima of the <scalar potential> have $|\phi|=v$. Choose the <vacuum expectation value> $\phi_0=ve_3$. Its <stabilizer subgroup> consists exactly of matrices $\operatorname{diag}(R_2,1)$ with $R_2\in O(2)$, so
$$
\boxed{O(3)\longrightarrow O(2).}
$$
The continuous <spontaneous symmetry breaking> is $SO(3)\to SO(2)$ and has two broken generators. For the full orthogonal group, the cross-product notation requires the <gauge connection> to transform in the adjoint, as an axial vector: $R[A]_\times R^{-1}=[(\det R)RA]_\times$. The <scalar field> transforms as the ordinary vector $R\phi$. This <axial transformation of orthogonal gauge connections> makes the stated <covariant derivative> transform correctly even for $\det R=-1$; assigning an ordinary-vector transformation to the connection would fail.
Near this nonzero vacuum, <unitary gauge> removes the two angular <Goldstone bosons> and gives $\phi=(0,0,v+h)$. Put
$$
W_\mu^\pm=\frac{A_\mu^1\mp iA_\mu^2}{\sqrt2},\qquad A_\mu=A_\mu^3,\qquad f_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.
$$
These are the physical charged <gauge bosons>, neutral massless <gauge boson>, and radial <scalar field>. Directly substituting into the original <gauge field strengths> gives
$$
\mathcal F_{\mu\nu}^\pm=(\partial_\mu\mp ieA_\mu)W_\nu^\pm-(\partial_\nu\mp ieA_\nu)W_\mu^\pm,\qquad \mathcal F_{\mu\nu}^3=f_{\mu\nu}-ie(W_\mu^+W_\nu^--W_\nu^+W_\mu^-).
$$
The <scalar field> <kinetic term> has no radial-vector cross term, because $\partial_\mu\phi$ and $\mathbf A_\mu\times\phi$ lie in perpendicular internal directions. The complete physical gauge-scalar <Lagrangian> becomes
$$
\mathcal L=-\frac14\mathcal F^{3\mu\nu}\mathcal F^3_{\mu\nu}-\frac12\mathcal F^{+\mu\nu}\mathcal F^-_{\mu\nu}+\frac12(\partial h)^2+e^2(v+h)^2W_\mu^+W^{-\mu}-\frac12\lambda v^2h^2-\frac12\lambda vh^3-\frac18\lambda h^4.
$$
For each real vector the <mass term> is $m^2A_\mu A^\mu/2$, while for the complex pair it is $m_W^2W_\mu^+W^{-\mu}$. Hence the <adjoint triplet Higgs spectrum> is
$$
\boxed{m_{W^+}=m_{W^-}=ev,\qquad m_A=0.}
$$
The radial scalar also has $m_h^2=\lambda v^2$ in the question's potential normalization. The two lost scalar modes supply the longitudinal polarizations of the massive vectors through the <Higgs mechanism>.
For the <lepton> <kinetic terms> let $P_L=(1-\gamma^5)/2$, $P_R=(1+\gamma^5)/2$, and denote the left neutral triplet entry by $n_L=c_\alpha\nu_L+s_\alpha N_L$. The missing singlet is the perpendicular combination
$$
\boxed{L^S=P_L(-\nu_e\sin\alpha+N\cos\alpha).}
$$
Its overall sign is immaterial. The real <orthogonal matrix> $\left(\begin{smallmatrix}c_\alpha&s_\alpha\\-s_\alpha&c_\alpha\end{smallmatrix}\right)$ is orthogonal, so the sum of the kinetic terms of $n_L$ and $L^S$ is $\bar\nu_Li\not\partial\nu_L+\bar N_Li\not\partial N_L$ with no mixed terms. Combining with the right triplet gives ordinary <Dirac field> <kinetic terms> for $E^+,N,e$, and only a left-handed <kinetic term> for the massless <neutrino>. This is <orthogonal completion of mixed neutral lepton kinetic terms>. The heavy charged field is $E^+$, as printed in the PDF, not the TeX's $N^+$.
To compute the charge commutator, the <canonical anticommutation relations> imply
$$
[\psi_i^\dagger(\mathbf x)\psi_j(\mathbf x),\psi_k^\dagger(\mathbf y)\psi_l(\mathbf y)]=\delta^3(\mathbf x-\mathbf y)(\delta_{jk}\psi_i^\dagger\psi_l-\delta_{il}\psi_k^\dagger\psi_j).
$$
The quartic terms cancel on reordering the fermions. Thus well-defined normal-ordered integrated bilinears obey <fermionic bilinear charge algebra>, $[Q_C,Q_D]=Q_{[C,D]}$. Left/right mixed <commutators> vanish since $P_LP_R=0$.
In either triplet, write
$$
K=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},\qquad [K,K^\dagger]=\operatorname{diag}(1,0,-1).
$$
The displayed current has a factor two relative to the projected bilinears, since $1\mp\gamma^5=2P_{L,R}$. Consequently its charges are $T^+=2(Q_K^L+Q_K^R)$ and $T^-=(T^+)^\dagger$. The neutral combinations have canonical anticommutators because $c_\alpha^2+s_\alpha^2=1$. Applying the matrix commutator gives
$$
[T^+,T^-]=4\int d^3x:(E^{+\dagger}E^+-e^\dagger e):=\boxed{\frac4eQ_\ell},
$$
where $Q_\ell=e\int:(E^{+\dagger}E^+-e^\dagger e):$ is the electromagnetic charge in the lepton sector. The neutrino, heavy neutral lepton and singlet contribute zero. This establishes <triplet charged-current closure on electromagnetic charge>, with the requested proportionality independent of normalization choices for the weak <Noether charges>. The PDF correctly writes $J^-=(J^+)^\dagger$; the TeX repeats $J^+$ incorrectly. Equal-time algebra does not require conservation of the separate fermionic weak currents after symmetry breaking. For full gauge-theory charges, the corresponding charged-boson contributions must also be included.
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