= Solution
{c}
For any fixed $r\geq0$, the source contribution obeys $c\phi_1\geq cr$, because $c<0$ and $\phi_1\leq r$. Equality occurs only at $(\phi_1,\phi_2,\phi_3)=(r,0,0)$ when $r>0$. Thus the global <ground state> lies on the positive first axis. Minimize there:
$$
f(r)=-\frac12\mu^2r^2+\frac14\lambda r^4+cr,\qquad
f'(r)=\lambda r^3-\mu^2r+c.
$$
For $0\leq r\leq v=\mu/\sqrt\lambda$, $f'(r)<0$. For $r>v$, $f''(r)=3\lambda r^2-\mu^2>0$, so $f'$ increases strictly from $c$ to $+\infty$. There is exactly one root $\tilde v>v$, and it is the unique global minimum. Therefore
$$
\boxed{\langle\boldsymbol\phi\rangle=(\tilde v,0,0),\qquad
\lambda\tilde v^3-\mu^2\tilde v+c=0,\qquad \tilde v>\frac\mu{\sqrt\lambda}>0.}
$$
This argument selects the true <ground state> even if the full cubic has additional negative stationary roots.
Expand $\phi_1=\tilde v+h$, $\phi_2=\pi_2$, $\phi_3=\pi_3$. The source is linear, so it shifts the expansion point but has zero <Hessian matrix>. The <scalar mass matrix> is diagonal at the new <vacuum expectation value>:
$$
m_h^2=-\mu^2+3\lambda\tilde v^2,\qquad
m_{\pi_2}^2=m_{\pi_3}^2=-\mu^2+\lambda\tilde v^2.
$$
Eliminating $\lambda$ with the stationary equation gives
$$
\boxed{m_h^2=2\mu^2-\frac{3c}{\tilde v},\qquad
m_{\pi_2}^2=m_{\pi_3}^2=-\frac c{\tilde v}.}
$$
All three squared <masses> are positive. The <masses> themselves are the positive square roots of these expressions.
The source explicitly reduces the stated <SO(3)> <symmetry> to <SO(2)>; the selected <vacuum state> preserves that remaining <SO(2)>. There is consequently no spontaneously broken exact continuous <global symmetry in field theory> to which the <Goldstone theorem> applies. For small $|c|$, the two angular excitations are <pseudo-Goldstone bosons>. Their squared <masses> vanish as $c\to0^-$, while $\tilde v\to\mu/\sqrt\lambda$ and $m_h^2\to2\mu^2$.
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