If the centroid of ∆ABC, which has vertices A(a, b), B(b, c) and C(c, a),

The given points are A(a, b), B(b, c) and C(c, a).
Here,

$\left(x_{1}=a, y_{1}=b\right), \quad\left(x_{2}=b, y_{2}=c\right)$ and $\left(x_{3}=c, y_{3}=a\right)$

Let the centroid be (x, y).
Then,

$x=\frac{1}{3}\left(x_{1}+x_{2}+x_{3}\right)$

$=\frac{1}{3}(a+b+c)$

$=\frac{a+b+c}{3}$

$y=\frac{1}{3}\left(y_{1}+y_{2}+y_{3}\right)$

$=\frac{1}{3}(b+c+a)$

$=\frac{a+b+c}{3}$

But it is given that the centroid of the triangle is the origin.
Then, we have:

$\frac{a+b+c}{3}=0$

$\Rightarrow a+b+c=0$