Explanation

(a)
Let $A \equiv \left| \begin{matrix}
a - x & c & b \\
c & b - x & a \\
b & a & c - x
\end{matrix} \right| = 0$
Applying $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$
$= \left| \begin{matrix}
a + b + c - x & c & b \\
a + b + c - x & b - x & a \\
a + b + c - x & a & c - x
\end{matrix} \right|$
$= (a + b + c - x)\left| \begin{matrix}
1 & c & b \\
1 & b - x & a \\
1 & a & c - x
\end{matrix} \right|$
$\Rightarrow \ (a + b + c - x)\left\lbrack 1\left\{ (b - x)(c - x) - a^{2} \right\} - c(c - x - a) + b(a - b + x) \right\rbrack = 0$
$\Rightarrow \ (a + b + c - x)\left\lbrack bc - bx - cx + x^{2} - a^{2} - c^{2} + xc + ac + ab - b^{2} + bx \right\rbrack = 0$
$\Rightarrow \ (a + b + c - x)\left\lbrack x^{2} - \left( a^{2} + b^{2} + c^{2} \right) + ab + bc + ca \right\rbrack = 0$
$\because\ ab + bc + ca = 0$ (given)
$\Rightarrow$ either $x = a + b + c\ or\ x = \left( a^{2} + b^{2} + c^{2} \right)^{1/2}$