= Solution
Take the <basis> $(H_1,H_2,E_{12},E_{21},E_{13},E_{31},E_{23},E_{32})$ of $\mathfrak{sl}_3$, where $H_1=\operatorname{diag}(1,-1,0)$, $H_2=\operatorname{diag}(0,1,-1)$ and $E_{ij}$ is a matrix unit. For $D=\operatorname{diag}(1,-2,1)$, $[D,H_r]=0$ and $[D,E_{ij}]=(d_i-d_j)E_{ij}$. Thus the <Adjoint representation> matrix is
$$
\operatorname{ad}D=\operatorname{diag}(0,0,3,-3,0,0,-3,3).
$$
Squaring and taking the <trace> computes the <Killing form> directly:
$$
\boxed{B(D,D)=\operatorname{tr}((\operatorname{ad}D)^2)=4\cdot9=36.}
$$
The independent identity $B(X,Y)=6\operatorname{tr}(XY)$ for the <special linear Lie algebra> $\mathfrak{sl}_3$ also gives $6(1+4+1)=36$.
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