System of equation x + 3y + 2z = 6 x+λy+2z=7x+3y+2z=μ has

x+3y+2z=6----ix+λy+2z=7----iix+3y+2z=μ----iii A) If λ=2, then D=0, therefore unique solution is not possible  B) If λ=4,μ=6x+3y=6−2zx+4y=7−2zy=1 and x=3−2zSubstituting in equation (iii) 3 – 2z + 3 + 2z = 6 is satisfied   Infinite solutions C) λ=5,μ=7 Consider equation (ii) and (iii) x+5y=7−2zx+3y=7−2z∴y=0x=7−2z are solution  Sub. In (i) 7−2z+2z=6 does not satisfy  no solution  D) if λ=3, μ=5then equation (i) and (ii) have no solution no solution