Determinants — Class 12 Maths Solution

Consider $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&{b + c}&1\\b&{c + a}&1\\c&{a + b}&1\end{array}} \right|$

Applying ${C_1} \to {C_1} + {C_2},$

we get
$= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{a + b + c}&{b + c}&1\\{a + b + c}&{c + a}&1\\{a + b + c}&{a + b}&1\end{array}} \right|$

Taking $(a + b + c)$ common from ${C_1}$,

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

$= (a + b + c) \times 0$
$=$ 0.

Hence, the points are collinear.