= Solution
{c}
Put $r^2=\Phi^T\Phi$. The potential has its minimum at
$$
\frac{dV}{dr}=-m^2r+4\lambda r^3=0,\qquad \boxed{v^2=\frac{m^2}{4\lambda}}.
$$
Choose the <vacuum expectation value> $\Phi_0=(0,0,v)^T$. Since the <weak hypercharge> is zero, $Q=T^3$ on this <real scalar triplet>, and $T^3\Phi_0=0$. Neither $T^1$ nor $T^2$ annihilates $\Phi_0$. The <stabilizer subgroup> is consequently
$$
\boxed{SU(2)\times U(1)_Y\longrightarrow U(1)_{T^3}\times U(1)_Y.}
$$
In particular, <spontaneous symmetry breaking> has not reduced the two neutral <gauge generators> to a single one: $Y$ annihilates every component and remains unbroken independently of $T^3$.
The <scalar mass matrix> is
$$
\left.\frac{\partial^2V}{\partial\phi_i\partial\phi_j}\right|_{\Phi_0}
=(-m^2+4\lambda v^2)\delta_{ij}+8\lambda(\Phi_0)_i(\Phi_0)_j
=8\lambda v^2\delta_{i3}\delta_{j3}.
$$
With canonical kinetic normalization $\tfrac12(\partial h)^2$, the radial <Higgs boson> therefore has
$$
\boxed{m_h^2=8\lambda v^2=2m^2,\qquad m_h=\sqrt2\,m.}
$$
The two tangential zero <eigenvalues> are <Goldstone modes>, absorbed by the two massive <gauge bosons> through the <Higgs mechanism>.
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