If $\mathrm{a}+\mathrm{b}+\mathrm{c}=0,$ and $\left|\begin{array}{ccc}\mathrm{a}-\mathrm{x}& \mathrm{c}& \mathrm{b}\\ \mathrm{c}& \mathrm{b}-\mathrm{x}& \mathrm{a}\\ \mathrm{b}& \mathrm{a}& \mathrm{c}-\mathrm{x}\end{array}\right|=0$

$$\begin{gathered} =\left|\begin{array}{ccc}\mathrm{a}-\mathrm{x}& \mathrm{c}& \mathrm{b}\\ \mathrm{c}& \mathrm{b}-\mathrm{x}& \mathrm{a}\\ \mathrm{b}& \mathrm{a}& \mathrm{c}-\mathrm{x}\end{array}\right| \\ \begin{array}{rcl}{\mathrm{R}}_1& \to & {\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3\end{array} \end{gathered}$$

$$\begin{gathered} \begin{array}{rcl} & \Rightarrow & \end{array}\begin{array}{rcl} & & \left|\begin{array}{ccc}-\mathrm{x}& -\mathrm{x}& -\mathrm{x}\\ \mathrm{c}& \mathrm{b}-\mathrm{x}& \mathrm{a}\\ \mathrm{b}& \mathrm{a}& \mathrm{c}-\mathrm{x}\end{array}\right|\end{array}\begin{array}{rcl} & =& \end{array}\begin{array}{rcl} & & 0\end{array}\begin{array}{rcl} & & \end{array}\begin{array}{rcl} & & \end{array}\begin{array}{rcl} & & \end{array}\begin{array}{rcl} & & \left(\therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=0\right)\end{array} \\ \begin{array}{rcl} & & \end{array}\begin{array}{rcl} & \Rightarrow & \end{array}\begin{array}{rcl} & & -\end{array}\begin{array}{rcl} & & \mathrm{x}\end{array}\begin{array}{rcl} & & \end{array}\begin{array}{rcl} & & \left|\begin{array}{ccc}1& 1& 1\\ \mathrm{c}& \mathrm{b}-\mathrm{x}& \mathrm{a}\\ \mathrm{b}& \mathrm{a}& \mathrm{c}-\mathrm{x}\end{array}\right|\end{array}\begin{array}{rcl} & =& \end{array}\begin{array}{rcl}0 {\mathrm{C}}_2& \to & {\mathrm{C}}_2-{\mathrm{C}}_1\\ {\mathrm{C}}_3& \to & {\mathrm{C}}_3-{\mathrm{C}}_1 \end{array}\begin{array}{rcl} & & \end{array} \\ \begin{array}{rcl} & & \end{array}\begin{array}{rcl} & & \mathrm{x}\end{array}\begin{array}{rcl} & =& \end{array}\begin{array}{rcl}0 \left(\mathrm{or}\right) \left|\begin{array}{ccc}1& 0& 0\\ \mathrm{c}& \mathrm{b}-\mathrm{x}-\mathrm{c}& \mathrm{a}-\mathrm{c}\\ \mathrm{b}& \mathrm{a}-\mathrm{b}& \mathrm{c}-\mathrm{x}-\mathrm{b}\end{array}\right|& =& 0\end{array}\begin{array}{rcl} & & \end{array} \end{gathered}$$

$\begin{array}{rcl}(\mathrm{b}-\mathrm{x}-\mathrm{c}) (\mathrm{c}-\mathrm{x}-\mathrm{b})-(\mathrm{a}-\mathrm{c}) (\mathrm{a}-\mathrm{b})& =& 0\\ \mathrm{bc}-\cancel{\mathrm{bx}}-{\mathrm{b}}^2-\cancel{\mathrm{cx}}+{\mathrm{x}}^2+\cancel{\mathrm{bx}}-{\mathrm{c}}^2+\cancel{\mathrm{cx}}+\cancel{\mathrm{bc}}-{\mathrm{a}}^2+\mathrm{ab}+\mathrm{ac}-\cancel{\mathrm{bc}}& =& 0\\ {\mathrm{x}}^2& =& {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2-(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})\\ \mathrm{wkt}, {(\mathrm{a}+\mathrm{b}+\mathrm{c})}^2& =& {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+2(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})\\ & \Rightarrow & \mathrm{ab}+\mathrm{bc}+\mathrm{ca}=-\frac{1}{2}\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)\\ \therefore {\mathrm{x}}^2& =& \frac{3}{2}\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)\\ \mathrm{x}& =& \pm \sqrt{\frac{3}{2} \left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)}\end{array}$