Show that the points $\left(a, b + c\right)$, $\left(b, c + a\right)$ and $\left(c, a + b\right)$ are collinear.
Let $A \left(x_1, y_1\right) = \left(a, b + c\right)$, $\;$ $B \left(x_2, y_2\right) = \left(b, c + a\right)$ and $C \left(x_3, y_3\right) = \left(c, a + b\right)$
Area of triangle formed by the points $ABC$ is
$\begin{aligned}
\Delta & = \dfrac{1}{2} \left[x_1 \left(y_2 - y_3\right) + x_2 \left(y_3 - y_1\right) + x_3 \left(y_1 - y_2\right)\right] \\\\
& = \dfrac{1}{2} \left[a \left(c + a - a - b\right) + b \left(a + b - b - c\right) + c \left(b + c -c - a\right)\right] \\\\
& = \dfrac{1}{2} \left[a \left(c - b\right) + b \left(a - c\right) + c \left(b - a\right)\right] \\\\
& = \dfrac{1}{2} \left[ac - ab + ab - bc + bc - ac\right] \\\\
& = 0
\end{aligned}$
$\because \;$ The area of triangle formed by the points $A$, $B$ and $C$ is zero,
$\implies$ the points $A$, $B$ and $C$ are collinear.