The area of the triangle formed by

We have three non-collinear points $\mathrm{A}(a, b+c) ; \mathrm{B}(b, c+a) ; \mathrm{C}(c, a+b)$.

In general if $\mathrm{A}\left(x_{1}, y_{1}\right) ; \mathrm{B}\left(x_{2}, y_{2}\right) ; \mathrm{C}\left(x_{3}, y_{3}\right)$ are non-collinear points then are of the triangle formed is given by-

$\operatorname{ar}(\Delta \mathrm{ABC})=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

So,

$a r(\Delta \mathrm{ABC})=\frac{1}{2}|a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)|$

$=\frac{1}{2}[a(c-b)+b(a-c)+c(b-a)]$

$=0$

So the answer is (d)