 Solution

Use a Hermitian gauge field $A_{\mu }=A_{\mu }^A{T}^A$ and $\mathrm{tr}(T^A{T}^B)={\delta }^{AB}/2$. With fundamental covariant derivative ${\partial }_{\mu }-ig{A}_{\mu }$, the adjoint derivative isIt transforms as $D_{\mu }\Phi \mapsto U(D_{\mu }\Phi )U^{†}$, provided $A_{\mu }^{\prime }=U{A}_{\mu }{U}^{†}-(i/g)({\partial }_{\mu }U)U^{†}$.

The infinitesimal parameter in $U=1+iϵ+O({ϵ}^2)$ must be Hermitian and traceless, from unitarity and unit determinant. It is not generally antisymmetric as the hint states. The valid generator $ϵ=\mathrm{diag}(1,-1,0,0,0)$ is a concrete counterexample to that assertion.

Choose canonical adjoint kinetic normalization ${\mathcal{L}}_{\Phi }=\mathrm{tr}(D_{\mu }\Phi D^{\mu }\Phi )$, which gives $\frac{1}{2}\partial {\Phi }^A\partial {\Phi }^A$. For the diagonal vacuum expectation value $D=v\mathrm{diag}(2,2,2,-3,-3)$, the matrix unit $E_{ij}$ obeys $[D,E_{ij}]=(d_i-d_j)E_{ij}$. Thus the mass of the gauge boson associated with either normalized Hermitian generator joining $i,j$ is $g|d_i-d_j|$.

Within the three-dimensional and two-dimensional blocks, or for diagonal generators, the commutator vanishes. There are therefore $8+3+1=12$ massless gauge bosons. The $3\times 2$ interblock pairs supply two real generators each, hence 12 equally massive gauge bosons:The absolute mass formula assumes the stated kinetic normalization, which the question leaves implicit. Writing instead $\frac{1}{2}\mathrm{tr}(D\Phi {)}^2$ without rescaling $\Phi$ would multiply these squared masses by $1/2$.

This is the gauge group of the Standard Model, up to its global quotient, arising from one adjoint Higgs mechanism in a simple unified group. The unbroken abelian generator can be chosen as $Y=\mathrm{diag}(-1/3,-1/3,-1/3,1/2,1/2)$, so $D=-6vY$. This is the characteristic first breaking step of an SU(5) grand unified theory, rather than electroweak breaking all the way to electromagnetism.