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Area of ΔABC is given by the relation,

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

$=\frac{1}{2}\left|\begin{array}{ccc}a & b+c & 1 \\ b-a & a-b & 0 \\ c-a & a-c & 0\end{array}\right|$ (Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$ )

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

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

$=0 \quad$ (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.