Given:
x+y+z=5
x+2y+λ2z=9
x+3y+λz=μ
⇒△=∣1111231λ2λ∣
⇒△=2λ−3λ2−λ+λ2+1
⇒△=−2λ2+λ+1
We know that for infinite solutions, \triangle =0&{\triangle }_{1}={\triangle }_{2}={\triangle }_{3}=0
⇒−2λ2+λ+1=0
⇒2λ2−λ−1=0
⇒(λ−1)(2λ+1)=0
⇒λ=1,2−1
Now, finding △1 for λ=1 we get,
Δ1=∣59μ123111∣
⇒Δ1=μ−13
And for λ=2−1 we get,
Δ1=∣59μ1231412−1∣=41∣59μ12341−2∣=47(13−μ)
So, for \lambda \neq 1&\mu \neq 13 system has no solution,
Hence, option (C) is the correct answer.