Q.Using determinants, show that the points $(a,b+c),(b,c+a),(c,a+b)$ are collinear.

[!TLDR] Apply $C_2\to C_2+C_1$: the new column 2 becomes a constant multiple of column 3, forcing the determinant to $0$. [!ANSWER] T …

The collinearity determinant is $\begin{vmatrix}a&b+c&1\\b&c+a&1\\c&a+b&1\end{vmatrix}$. Apply $C_2\to C_2+C_1$ (Property 6, unchanged value): the new column 2 becomes $(b+c+a,\ c+a+b,\ a+b+c)=(a+b+c,\,a+b+c,\,a+b+c)$, a constant column. This new column 2 is exactly $(a+b+c)$ times column 3 $=(1,1,1)$, so columns 2 and 3 are proportional; taking the factor $(a+b+c)$ out of column 2 (Property 4) leaves two identical columns, giving determ …

Set up the $3\times3$ area determinant for the three given points, apply $C_2\to C_2+C_1$ to reveal that the resulting columns are proport …

Trying to expand the determinant directly in terms of $a,b,c$ without first simplifying via a column operation -- this works but is far more …