Determinants — Class 12 Maths Solution

We have
$\left| {\begin{array}{llllllllllllllllllll}{a - b}&{b + c}&a\\{b - a}&{c + a}&b\\{c - a}&{a + b}&c\end{array}} \right|$

$= \left| {\begin{array}{llllllllllllllllllll}{a + c}&{b + c + a}&a\\{b + c}&{c + a + b}&b\\{c + b}&{a + b + c}&c\end{array}} \right|$ and $\left. {{C_2} \to {C_2} + {C_3}} \right]$

$= (a + b + c)\left| {\begin{array}{llllllllllllllllllll}{a + c}&1&a\\{b + c}&1&b\\{c + b}&1&c\end{array}} \right|$

[taking $(a + b + c)$ common from $\left. {{C_2}} \right]$

$= (a + b + c)\left| {\begin{array}{cccccccccccccccccccc}{a - b}&0&{a - c}\\0&0&{b - c}\\{c + b}&1&c\end{array}} \right|$ [and ${R_1} \to {R_1} - {R_3}$]
$= (a + b + c)[ - (b - c) \cdot (a - b)]$

[expanding along ${R_2}$]
$= (a + b + c)(c - b)(a - b)$