We know that, the area of triangle with three collinear points is zero.
Now, consider Area of △ABC=21∣∣abcb+cc+aa+b111∣∣ =21∣a{(c+a)×1−(a+b)×1}−(b+c){b×1−1×c} +1{b×(a+b)−(c+a)×c}∣ =21∣a(c+a−a−b)−(b+c)(b−c) +1(ab+b2−c2−ac)∣ =21∣∣ac−ab−b2+c2+ab+b2−c2−ac∣∣=21×0=0
Since, area of △ABC=0.
Hence, points A(a,b+c),B(b,c+a) and C(c,a+b) are collinear.