⇒33−2(γ−22)+7(−3)=0γ=28Δ1=21α−2(11β−αγ)−33=21α−22β+2αγ−33Δ2=11β−αγ−7(β−2α)+γ−22=14α+4β+γ−αγ−22(P) If β=21(7α−3) and γ=28Δ=0,Δ1=0,Δ2=0,Δ3=0Infinitely many solutions x=11,y=−2 and z=0 will satisfy all the three given equations, so it is a solution. (Q) If β=21(7α−3) and γ=28 then Δ=0, but Δ3=0 so no solution(R) If β=21(7α−3),α=1 and γ=28Δ=0,Δ3=0 so a unique solution (S) If β=21(7α−3),α=1,γ=28Δ=0,Δ3=0,Δ1=0,Δ2=0, so a unique solution x=11,y=−2 and z=0 will satisfy all the three equations Option A is correct.