If vectors $\overline{\mathrm{a}}, \overline{\mathrm{b}}\text{ and }\overline{\mathrm{c}}$ are non-coplanar and $\mathcal{l}, \mathrm{m} \mathrm{and} \mathrm{n}$ are distinct scalars, then $\left[\mathcal{l}\overline{a}+m\overline{b}+n\overline{c} \mathcal{l}\overline{b}+m\overline{c}+n\overline{a} \mathcal{l}\overline{c}+m\overline{a}+n\overline{b}\right]=0$ implies

$\begin{array}{l}[\mathcal{l}\overline{a}+m\overline{b}+n\overline{c} \mathcal{l}\overline{b}+m\overline{c}+n\overline{a} \mathcal{l}\overline{c}+m\overline{a}+n\overline{b}]=0\\ \Rightarrow \left|\begin{array}{ccc}\mathcal{l}& m& n\\ n& \mathcal{l}& m\\ m& n& \mathcal{l}\end{array}\right|=0\end{array}$
$$\begin{gathered} (\mathcal{l}+m+n)(\mathcal{l}-m{)}^2+(m-n{)}^2+(n-1{)}^2=0 \\ \mathcal{l}+m+n=0 \\ \text{ Obviously, }\mathcal{\text{l}}{\mathrm{x}}^2+\mathrm{mx}+\mathrm{n}=0\text{ is satisfied by }\mathrm{x}=1 \end{gathered}$$