The area of the triangle with vertices [(a+1)(a+2),(a+2)],[(a+2)(a+3),(a+3)]  and [(a+3)(a+4),(a+4)]  is

Δ=12(a+1)(a+2)a+21(a+2)(a+3)a+31(a+3)(a+4)a+41 Applying R2→R2−R1  and R3→R3−R2,  we getΔ=12(a+1)(a+2)a+212(a+2)102(a+3)10=|−1|=1Which is independent of a.