If the system of equations x+y+z=0,x+2y+3z=1,λx+μy+4z=1 has infinitely many solutions then (λ,μ)=
(2, 3)
(3, 2)
(1, 2)
(2, 1)
x+y+z=0,………(1)x+2y+3z=1……… (2) λx+μy+4z=1…… (3) 3×(1)-(2) gives 2x+y=−1………...(4)4×(1)-(3) gives (4−λ)x+(4−μ)y=−1… (5) (4), (5) have infinitely many solutions ⇒24−λ=14−μ=1⇒λ=2,μ=3