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

(N/A) The area of $\triangle ABC$ is given by the determinant formula:
$\Delta = \frac{1}{2} \left| \begin{array}{lll} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array} \right|$
Applying the row operations $R_{2} \rightarrow R_{2} - R_{1}$ and $R_{3} \rightarrow R_{3} - R_{1}$:
$\Delta = \frac{1}{2} \left| \begin{array}{ccc} a & b+c & 1 \\ b-a & a-b & 0 \\ c-a & a-c & 0 \end{array} \right|$
Taking $(b-a)$ common from $R_{2}$ and $(c-a)$ common from $R_{3}$:
$\Delta = \frac{1}{2} (b-a)(c-a) \left| \begin{array}{ccc} a & b+c & 1 \\ -1 & 1 & 0 \\ 1 & -1 & 0 \end{array} \right|$
Applying $R_{3} \rightarrow R_{3} + R_{2}$:
$\Delta = \frac{1}{2} (b-a)(c-a) \left| \begin{array}{ccc} a & b+c & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right|$
Since all elements of the third row are $0$,the value of the determinant is $0$.
Thus,the area of the triangle formed by points $A, B$,and $C$ is $0$.
Hence,the points $A, B$,and $C$ are collinear.