Consider the system of linear equations x+y+z=5,x+2y+λ ^2z=9 and x+3y+λ z=μ, where λ ,μ ∈ R. Then, which of the following statement is NOT correct ?

Given:

$$x+y+z=5$$x+y+z=5

$$x+2y+{\lambda }^{2}z=9$$x+2y+λ2z=9

$$x+3y+\lambda z=\mu$$x+3y+λz=μ

$$\Rightarrow \triangle =|\begin{matrix}1 & 1 & 1 \\ 1 & 2 & {\lambda }^{2} \\ 1 & 3 & \lambda \end{matrix}|$$⇒△=∣111​123​1λ2λ​∣

$$\Rightarrow \triangle =2\lambda -3{\lambda }^{2}-\lambda +{\lambda }^{2}+1$$⇒△=2λ−3λ2−λ+λ2+1

$$\Rightarrow \triangle =-2{\lambda }^{2}+\lambda +1$$⇒△=−2λ2+λ+1

We know that for infinite solutions, \triangle =0&{\triangle }_{1}={\triangle }_{2}={\triangle }_{3}=0

$$\Rightarrow -2{\lambda }^{2}+\lambda +1=0$$⇒−2λ2+λ+1=0

$$\Rightarrow 2{\lambda }^{2}-\lambda -1=0$$⇒2λ2−λ−1=0

$$\Rightarrow (\lambda -1)(2\lambda +1)=0$$⇒(λ−1)(2λ+1)=0

$$\Rightarrow \lambda =1,\frac{-1}{2}$$⇒λ=1,2−1​

Now, finding ${\triangle }_{1}$△1​ for $\lambda =1$λ=1 we get,

$${\Delta }_{1}=|\begin{matrix}5 & 1 & 1 \\ 9 & 2 & 1 \\ \mu & 3 & 1\end{matrix}|$$Δ1​=∣59μ​123​111​∣

$$\Rightarrow {\Delta }_{1}=\mu -13$$⇒Δ1​=μ−13

And for $\lambda =\frac{-1}{2}$λ=2−1​ we get,

$${\Delta }_{1}=|\begin{matrix}5 & 1 & 1 \\ 9 & 2 & \frac{1}{4} \\ \mu & 3 & \frac{-1}{2}\end{matrix}|=\frac{1}{4}|\begin{matrix}5 & 1 & 4 \\ 9 & 2 & 1 \\ \mu & 3 & -2\end{matrix}|=\frac{7}{4}(13-\mu )$$Δ1​=∣59μ​123​141​2−1​​∣=41​∣59μ​123​41−2​∣=47​(13−μ)

So, for \lambda \neq 1&\mu \neq 13 system has no solution,

Hence, option $(C)$(C) is the correct answer.