The given points are collinear, if ∣∣t1t2t32at1+at132at2+at232at3+at33111∣∣=0 ⇒a∣∣t1t2t32t1+t132t2+t232t3+t33111∣∣=0
Applying R2→R2−R1,R3→R3−R1, we get ∣∣t1t2−t1t3−t12t1+t132(t2−t1)+(t23−t13)2(t3−t1)+(t33−t13)100∣∣ ⇒(t2−t1)(t3−t1) ∣∣t1112t1+t132+t22+t12+t2t12+t32+t12+t3t1100∣∣=0 ⇒(t2−t1)(t3−t1)(t3−t2)(t3+t2+t1)=0 ⇒t1+t2+t3=0[∵t1=t2=t3]