⇒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.
