= Solution
Parameterize the scalar locally as
$$
\Phi(x)=\exp\left[\frac{i\pi^\alpha(x)X^\alpha}{v}\right]
\frac1{\sqrt2}(0,0,v+h(x))^T,
$$
where the five $X^\alpha$ are broken generators. A gauge transformation removes all $\pi^\alpha$ in <unitary gauge>. Their derivative terms combine with the corresponding gauge fields in $(D_\mu\Phi)^\dagger D^\mu\Phi$, supplying the longitudinal polarizations of five massive vectors. This is the <Higgs mechanism>.
With $T^a=\lambda^a/2$, the gauge bosons $A_\mu^{1,2,3}$ of the unbroken $SU(2)$ remain massless. The four bosons $A_\mu^{4,5,6,7}$ have
$$
m_{4,5,6,7}^2=\frac{g^2v^2}{4},
$$
and $A_\mu^8$ has
$$
m_8^2=\frac{g^2v^2}{3}.
$$
The five eaten Goldstone modes complete their third polarizations. The sixth real component of the complex triplet remains as a radial <Higgs boson> with
$$
m_h^2=2\lambda v^2.
$$
This realizes the degree-of-freedom count summarized by <Fundamental-Higgs breaking of SU(3) to SU(2)>.
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