(b,c,d)

x + 3y + 2z = 6 ..... (i)

x + λ y + 2z = 7 ..... (ii)

x + 3y + 2z = μ ..... (iii)

If λ = 2, then D = 0, therefore unique solution is not possible

If λ = 4, μ = 6

x + 3y = 6 – 2z

x + 4y = 7 – 2z

∴ y = 1 and x = 3 – 2z

substituting in equation (iii)

3 – 2z + 3 + 2z = 6 is satisfied

∴ infinite solutions

λ = 5, μ = 7

consider equation (ii) and (iii)

x + 5y = 7 – 2z

x + 3y = 7 – 2z

∴ y = 0 x = 7 – 2z are solution

sub. in (i)

7 – 2z + 2z = 6 does not satisfy

∴ no solution

If λ = 3, μ = 5

then equation (i) and (ii) have no solution

∴ no solution