The determinant $\left| {\begin{array}{llllllllllllllllllll}{{b^2} - ab}&{b - c}&{bc - ac}\\{ab - {a^2}}&{a - b}&{{b^2} - ab}\\{bc - ac}&{c - a}&{ab - {a^2}}\end{array}} \right|$ equals to

We have
$\left| {\begin{array}{llllllllllllllllllll}{{b^2} - ab}&{b - c}&{bc - ac}\\{ab - {a^2}}&{a - b}&{{b^2} - ab}\\{bc - ac}&{c - a}&{ab - {a^2}}\end{array}} \right|$

$= \left| {\begin{array}{llllllllllllllllllll}{b(b - a)}&{b - c}&{c(b - a)}\\{a(b - a)}&{a - b}&{b(b - a)}\\{c(b - a)}&{c - a}&{a(b - a)}\end{array}} \right|$

$= {(b - a)^2}\left| {\begin{array}{llllllllllllllllllll}b&{b - c}&c\\a&{a - b}&b\\c&{c - a}&a\end{array}} \right|$

[on taking $(b - a)$ common from ${C_1}$

and ${C_3}$ each]$= {(b - a)^2}\left| {\begin{array}{llllllllllllllllllll}{b - c}&{b - c}&c\\{a - b}&{a - b}&b\\{c - a}&{c - a}&a\end{array}} \right|$

$= 0$
[since, two columns ${C_1}$ and ${C_2}$ are identical, so the value of determinant is zero]