= Solution
Evaluate the <Killing form> on the <root-space decomposition>. For arbitrary $K\in\mathfrak h$,
$$
\mathcal B(H_\alpha)(K)=B(H_\alpha,K)
=\sum_{\beta\in\Phi}\beta(H_\alpha)\beta(K).
$$
Thus the <Killing dual of a coroot> is the <root> expansion
$$
\boxed{\mathcal B(H_\alpha)=\sum_{\beta\in\Phi}\beta(H_\alpha)\beta
=2\sum_{\beta\in\Phi^+}\beta(H_\alpha)\beta.}
$$
The factor two in the positive-root expression occurs because the contributions of $\beta$ and $-\beta$ are equal.
For $\mathfrak{sl}_3$, let $a=L_1-L_2$, $b=L_2-L_3$; the <positive roots> are $a,b,a+b$. On $H_{12}$ their values are $2,-1,1$. Therefore
$$
\mathcal B(H_{12})=2[2a-b+(a+b)]=6a,
\qquad
\boxed{\mathcal B(H_{12})=6(L_1-L_2)=12\omega_1-6\omega_2.}
$$
This uses the unscaled <Killing form> $\operatorname{tr}(\operatorname{ad}A\operatorname{ad}C)$. Changing that normalization would change the scalar six.
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