$ \Rightarrow $ $\left| {\,\begin{array}{*{20}{c}}{a + b + c - x}&c&b\\{a + b + c - x}&{b - x}&a\\{a + b + c - x}&a&{c - x}\end{array}\,} \right| = 0$

$ \Rightarrow $ $(x - \sum a )\,\left| {\begin{array}{*{20}{c}}1&c&b\\1&{b - x}&a\\1&a&{c - x}\end{array}\,} \right| = 0$

$ \Rightarrow $ $x = \sum {a = 0} $ (by hypothesis)

or $1\,\{ (b - x)\,(c - x) - {a^2}\} - c\{ c - x - a\} + b\{ a - b + x\} = 0$ by expanding the determinant.

or ${x^2} - ({a^2} + {b^2} + {c^2}) + (ab + bc + ca) = 0$

or ${x^2} - \left( {\sum {{a^2}} } \right) - \frac{1}{2}\,\left( {\sum {{a^2}} } \right) = 0$

$\left\{ {\,a + b + c = 0 \Rightarrow {{(a + b + c)}^2} = 0} \right.$

$ \Rightarrow $ $\left. {\sum {{a^2}} + 2\sum {ab} = 0 \Rightarrow \sum {ab} = - \frac{1}{2}\sum {{a^2}} } \right\}$

or $x = \pm \sqrt {\frac{3}{2}\sum {{a^2}} } $

$\therefore $ The solution is $x = 0$ or $ \pm \sqrt {\frac{3}{2}\sum {{a^2}} } $.

Trick: Put $a = 1,\,b = - 1$ and $c = 0$ so that they satisfy the condition $a + b + c = 0$.

Now the determinant becomes $\left| {\,\begin{array}{*{20}{c}}{1 - x}&0&{ - 1}\\0&{ - 1 - x}&1\\{ - 1}&1&{ - x}\end{array}\,} \right| = 0$

$ \Rightarrow $ $(1 - x)\,\{ x(1 + x) - 1\} + 1(1 + x) = 0$

$ \Rightarrow $ $(1 - x)\,\{ {x^2} + x - 1\} + x + 1 = 0$

$ \Rightarrow $ $x({x^2} - 3) = 0$

Now putting these in the options, we find that option $ (c)$  gives the same values i.e., $0,  \pm \sqrt 3 $.