= Solution
A stable symmetry-breaking minimum requires $\lambda>0$ and $m^2<0$. Put $v^2=-m^2/\lambda>0$. The common <scalar potential> can then be written, up to a constant, as
$$
V=\frac\lambda4(\Phi\cdot\Phi-v^2)^2.
$$
Its minima form the sphere $\Phi\cdot\Phi=v^2$. Choose the vacuum direction $\Phi_0=(0,0,v)^T$. If instead $m^2>0$ with positive $\lambda$, the minimum is at the origin and the assumed nonzero <vacuum expectation value> is not a stable vacuum of these models.
In the ungauged theory, the internal symmetry is global $O(3)$; its connected part is $SO(3)$. Constant <orthogonal> transformations preserve both the <kinetic term> and the potential. The chosen vacuum is invariant under $O(2)$ acting on its first two components, so the continuous breaking is $SO(3)\to SO(2)$ with two broken generators. The vacuum orientations are genuinely different degenerate global vacua.
The <global triplet scalar symmetry breaking> is visible directly in the fluctuation masses. Write $\Phi=(\pi_1,\pi_2,v+h)^T$. The potential Hessian is
$$
\left.\frac{\partial^2V}{\partial\Phi_a\partial\Phi_b}\right|_{\Phi_0}
=(m^2+\lambda v^2)\delta_{ab}+2\lambda\Phi_{0a}\Phi_{0b}
=\operatorname{diag}(0,0,2\lambda v^2).
$$
The quadratic Lagrangian is therefore
$$
\mathcal L_1^{(2)}=\frac12(\partial h)^2+\frac12(\partial\pi_1)^2+\frac12(\partial\pi_2)^2-\frac12(2\lambda v^2)h^2.
$$
\b[There is one radial scalar of <mass> $m_h^2=2\lambda v^2=-2m^2$ and two massless <Goldstone bosons>.] The two Goldstone modes describe motion tangential to the vacuum sphere. They remain physical, since a global rotation has only constant parameters and cannot remove arbitrary spacetime-dependent fluctuations. All three original real-scalar degrees of freedom remain. Interactions persist: the shifted potential is $V-V_{\min}=\frac\lambda4(2vh+h^2+\pi_1^2+\pi_2^2)^2$, containing cubic and quartic terms. Masslessness of the angular modes is protected by the exact continuous symmetry through Goldstone's theorem.
In the gauged theory, the three-component real field is the vector representation of local $SO(3)$, equivalently the <Adjoint representation> of $SU(2)$. The center of $SU(2)$ acts trivially on this field, so both descriptions have the same local field content and perturbative spectrum. Take $(t^a)_{bc}=-i\epsilon^{abc}$, for which $[t^a,t^b]=i\epsilon^{abc}t^c$. The covariant <kinetic term> and the gauge <kinetic term> are invariant under local transformations with the associated connection transformation. The scalar inversion $\Phi\mapsto-\Phi$, with the <gauge field> unchanged, is also a discrete invariance of the displayed action.
The vacuum leaves rotations about its third axis unbroken. The residual gauge group is $SO(2)$, or the corresponding $U(1)$ subgroup in the $SU(2)$ description. Expanding the covariant <kinetic term> about $\Phi_0$ gives
$$
\frac12(D_\mu\Phi_0)\cdot(D^\mu\Phi_0)
=\frac12g^2v^2\left(B_\mu^1B^{1\mu}+B_\mu^2B^{2\mu}\right).
$$
Thus the <gauge-boson mass matrix> is $g^2v^2\operatorname{diag}(1,1,0)$. The third generator annihilates the vacuum, while the first two do not. \b[The masses are $m_{B^1}=m_{B^2}=gv$, $m_{B^3}=0$, and $m_h=\sqrt{2\lambda}\,v$.]
The quadratic derivative mixing displays what happens to the Goldstone fields. With the generator convention just chosen,
$$
(D_\mu\Phi)_1=\partial_\mu\pi_1-gvB_\mu^2+\cdots,\quad
(D_\mu\Phi)_2=\partial_\mu\pi_2+gvB_\mu^1+\cdots,\quad
(D_\mu\Phi)_3=\partial_\mu h+\cdots.
$$
Local rotations can set the two angular fields to zero in <unitary gauge>. Then $\Phi=(0,0,v+h)^T$ and
$$
\mathcal L_{\rm scalar}=\frac12(\partial h)^2
+\frac12g^2(v+h)^2\left[(B^1)^2+(B^2)^2\right]-V(v+h).
$$
The two Goldstone modes supply the <longitudinal polarizations> of the two massive vector bosons. They do not survive as additional physical massless scalars. The radial scalar is still physical, and $B^3$ retains only two <transverse polarizations>. This <adjoint triplet Higgs spectrum> realizes the <Higgs mechanism>; the gauge group is only partially broken, so one massless <gauge boson> remains.
The degree-of-freedom check is
$$
\boxed{\text{before: }3+3\times2=9,\qquad
\text{after: }1+2\times3+1\times2=9.}
$$
In contrast, the global theory has just the three scalar modes, including its two physical <Goldstone bosons>. A <vacuum expectation value> in the local theory is a choice of gauge-fixed description: the direction can be changed by <gauge transformations> and is not itself an observable. Gauge-independent masses and the physical degree count express the actual effect. Both theories retain Lorentz symmetry in the constant scalar vacuum.
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