= Solution
{c}
Put $r^2=\phi_i\phi_i$. Completing the square in the <scalar potential> gives
$$
V=\frac\lambda4\left(r^2-\frac{\mu^2}{\lambda}\right)^2-\frac{\mu^4}{4\lambda}.
$$
Since $\lambda>0$, its global minima form the <vacuum manifold>
$$
\boxed{r=v,\qquad v=\frac{\mu}{\sqrt\lambda}.}
$$
Here $\mu$ denotes the positive square root of the parameter $\mu^2$. The origin is a local maximum, not a <ground state>. Choose $\langle\boldsymbol\phi\rangle=(v,0,0)$. Its <stabilizer subgroup> is the <SO(2)> subgroup of rotations in the $(2,3)$ plane, so the stated <SO(3)> <global symmetry in field theory> is spontaneously broken to <SO(2)>.
Write $\phi_1=v+h$, $\phi_2=\pi_2$, $\phi_3=\pi_3$. The <kinetic terms> stay canonically normalized, while the <scalar mass matrix> is
$$
(M^2)_{ij}=(-\mu^2+\lambda\phi^2)\delta_{ij}+2\lambda\phi_i\phi_j,
\qquad
M^2\big|_{(v,0,0)}=\operatorname{diag}(2\mu^2,0,0).
$$
Equivalently, the quadratic <scalar potential> is $\mu^2h^2$. Therefore the particle <masses> are
$$
\boxed{m_h=\sqrt2\mu,\qquad m_{\pi_2}=m_{\pi_3}=0.}
$$
The two angular <Goldstone bosons> match $\dim SO(3)-\dim SO(2)=3-1=2$. The radial displacement changes $r$ and costs <potential energy>, giving the massive <Higgs mode>. These are tree-level <masses>; the <Goldstone theorem> protects the angular zero-<mass> modes in the exact global theory.
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