Matrices and Determinants

x+2y+z=7
x+αz=11
2x−3y+βz=γ
∆=|

1	2	1
1	0	α
2	−3	β

|=0
3α−2(β−2α)−3=0
7α−2β=3
⇒β=‌

1

2

(7α−3)
⇒33−2(γ−22)+7(−3)=0
γ‌=28
∆1‌=21α−2(11β−αγ)−33
‌=21α−22β+2αγ−33
∆2‌=11β−αγ−7(β−2α)+γ−22
‌=14α+4β+γ−αγ−22
(P) If β=‌

1

2

(7α−3) and γ=28
∆=0,∆1=0,∆2=0,∆3=0
Infinitely many solutions
x=11,y=−2 and z=0 will satisfy all the three given equations, so it is a solution.
(Q) If β=‌

1

2

(7α−3) and γ≠28 then ∆=0, but ∆3≠0 so no solution
(R) ‌ If ‌β≠‌

1

2

(7α−3),α=1‌ and ‌γ≠28
∆≠0,∆3≠0‌ so a unique solution ‌
(S) ‌ If ‌β≠‌

1

2

(7α−3),α=1,γ=28
∆≠0,∆3=0,∆1≠0,∆2≠0,‌ so a unique solution ‌
x=11,y=−2‌ and ‌z=0‌ will satisfy all the three equations ‌
Option A is correct.