If $f(x) = \left| {\begin{array}{cccccccccccccccccccc}0&{x - a}&{x - b}\\{x + a}&0&{x - c}\\{x + b}&{x + c}&0\end{array}} \right|$, then

We have, $f(x) = \left| {\begin{array}{cccccccccccccccccccc}0&{x - a}&{x - b}\\{x + a}&0&{x - c}\\{x + b}&{x + c}&0\end{array}} \right|$

$\Rightarrow$ $f(a) = \left| {\begin{array}{cccccccccccccccccccc}0&0&{a - b}\\{2a}&0&{a - c}\\{a + b}&{a + c}&0\end{array}} \right|$

$= [(a - b)\{ 2a \cdot (a + c)\} ] \ne 0$
$\therefore f(b) = \left| {\begin{array}{cccccccccccccccccccc}0&{b - a}&0\\{b + a}&0&{b - c}\\{2b}&{b + c}&0\end{array}} \right|$

$= - (b - a)[2b(b - c)]$
$= - 2b(b - a)(b - c) \ne 0$

$\therefore f(0) = \left| {\begin{array}{cccccccccccccccccccc}0&{ - a}&{ - b}\\a&0&{ - c}\\b&c&0\end{array}} \right|$ \[\]
$= a(bc) - b(ac)$

$= abc - abc = 0$