= Solution
In the ordered basis $(K_1,K_2,J)$, the three <Adjoint representation> matrices are
$$
A_{K_1}=\begin{pmatrix}0&0&0\\0&0&-\lambda\\0&1&0\end{pmatrix},\quad
A_{K_2}=\begin{pmatrix}0&0&\lambda\\0&0&0\\-1&0&0\end{pmatrix},\quad
A_J=\begin{pmatrix}0&-\lambda&0\\\lambda&0&0\\0&0&0\end{pmatrix}.
$$
Their cross-product traces vanish, while their squared traces are $-2\lambda$, $-2\lambda$ and $-2\lambda^2$. Consequently
$$
\boxed{\kappa=-2\operatorname{diag}(\lambda,\lambda,\lambda^2).}
$$
This is the <Killing form for cyclic three-generator brackets> with constants $(1,\lambda,\lambda)$.
For $\lambda>0$, the form is nondegenerate and negative-definite. The algebra is \b[semisimple and compact], isomorphic to the <SU(2) Lie algebra>. For $\lambda<0$, it is nondegenerate with signature $(2,1)$, so the algebra is \b[semisimple and noncompact], isomorphic to $\mathfrak{sl}_2(\mathbb R)$.
For $\lambda=0$, the only nonzero bracket is $[K_1,K_2]=J$, with $J$ central. This is the <Heisenberg Lie algebra>, which is nilpotent and \b[neither semisimple nor compact]. Compactness is ruled out even though the Killing form is merely degenerate: $A_{K_1}$ is nonzero and nilpotent, whereas in a <compact Lie algebra> every adjoint map is skew-adjoint for a positive inner product. <Skew-adjoint nilpotent operators vanish>.
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