Algebra + Geometry

We know that 3 points are colinear then are of triangle formed by them must be zero.

$\frac{1}{2} \begin{vmatrix} a^3+2a^2 & a^2 & 1 \\ b^3+2b^2 & b^2 & 1 \\ c^3+2c^2 & c^2 & 1 \end{vmatrix} = 0$

$\implies \begin{vmatrix} a^3 & a^2 & 1 \\ b^3 & b^2 & 1 \\ c^3 & c^2 & 1 \end{vmatrix} = 0$

$\iff \begin{vmatrix} a^3 & a^2 & 1 \\ b^3-a^3 & b^2-a^2 & 0 \\ c^3-a^3 & c^2-a^2 & 0 \end{vmatrix} = 0$

We can expand the determinant from here and get the result $ab+bc+ca=0$ . Note that while expanding or solving the determinant, we must keep in mind that $a \neq b \neq c$ .

Let the equation of line be $Ax+By+C=0$ , then by satisfying the given points in line give us the equation,

$Ax^3+(2A+B)x^2+C=0, \quad x=a,b,c$

By the above equation we can say that $ab+bc+ca=0$ .

PLOTTING MAKE IT MORE EASY TO REALIZE.