Determinant of $r^2\mathbb{1}-3\boldsymbol{r}\otimes\boldsymbol{r}$
\begin{align}
\boldsymbol{r}\otimes\boldsymbol{r}
=\begin{pmatrix}
x^2&xy&xz\\
yx&y^2&yz\\
zx&zy&z^2
\end{pmatrix}
\end{align}
This matrix has following properties:
- $\mathrm{rank}\,\boldsymbol{r}\otimes\boldsymbol{r}=1$
- $\mathrm{Tr}\,\boldsymbol{r}\otimes\boldsymbol{r}=r^2$
Therefore, we can change basis such that*1
\begin{align}
P^{-1}(\boldsymbol{r}\otimes\boldsymbol{r})P
=\begin{pmatrix}
r^2&0&0\\
*&0&0\\
*&0&0
\end{pmatrix}
\end{align}
Consequently (Note: $\det A=\det (P^{-1}AP)$),
\begin{align}
\det \left(r^2\mathbb{1}-3\boldsymbol{r}\otimes\boldsymbol{r}\right)
=-2r^6.
\end{align}
*1:Chose one from $\mathrm{Im}\,(\boldsymbol{r}\otimes\boldsymbol{r})$ and the others from $\mathrm{Ker}\,(\boldsymbol{r}\otimes\boldsymbol{r})$
