Show that points $A(a, b+c), B(b, c+a), C(c, a+b)$ are collinear

Area of $\triangle \mathrm{ABC}$ is given by the relation,

$\Delta=\frac{1}{2}\left|\begin{array}{lll}a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}a & b+c & 1 \\ b-a & a-b & 0 \\ c-a & a-c & 0\end{array}\right|$ 

( Applying $ R_{2} \rightarrow R_{2}-R_{1}$ and  $R_{3} \rightarrow R_{3}-R_{1}$)

$=\frac{1}{2}(a-b)(c-a)\left|\begin{array}{ccc}a & b+c & 1 \\ -1 & 1 & 0 \\ 1 & -1 & 0\end{array}\right|$

$=\frac{1}{2}(a-b)(c-a)\left|\begin{array}{ccc}a & b+c & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 0\end{array}\right| \quad\left(\text { Applying } R_{3} \rightarrow R_{3}+R_{2}\right)$

$=0 \quad$ (All elements of $R_{3}$ are $0$ )

Thus, the area of the triangle formed by points $A, B$ and $C$ is zero.

Hence, the points $A, B$ and $C$ are collinear.