Given system of equations x+y+az=2.....‌ (i) ‌ 3x+y+z=4.....‌ (ii) ‌ x+2z=1....‌ (iii) ‌ Solving (i), (ii) and (iii), we get x=1,y=1,z=0( and for unique solution a≠−3) Now, (α,1),(1,α) and (1,−1) are collinear ∴|
α
1
1
1
α
1
1
−1
1
|=0 ⇒α(α+1)−1(0)+1(−1−α)=0 ⇒α2−1=0 ∴α=±1 ∴ Sum of absolute values of α=1+1=2