 Solution

Evaluate the Killing form on the root-space decomposition. For arbitrary $K\in \mathfrak{h}$,Thus the Killing dual of a coroot is the root expansionThe 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 $2,-1,1$. ThereforeThis uses the unscaled Killing form $\mathrm{tr}(\mathrm{ad}A\mathrm{ad}C)$. Changing that normalization would change the scalar six.