Consider the system of linear equations x + y + z = 5, x + 2y + λ ^...

Given: 

$x + y + z = 5$

$x + 2y + {\lambda}^{2}z = 9$

$x + 3y + \lambdaz = \mu$

⇒ △ = \left|\begin{cases}1 & 1 & 1 \\ 1 & 2 & {\lambda}^{2} \\ 1 & 3 & \lambda\end{cases}\right|

$⇒ △ = 2\lambda - 3{\lambda}^{2} - \lambda + {\lambda}^{2} + 1$

$⇒ △ = - 2{\lambda}^{2} + \lambda + 1$

We know that for infinite solutions, $△ = 0 \text{ and } {△}_{1} = {△}_{2} = {△}_{3} = 0$

$⇒ - 2{\lambda}^{2} + \lambda + 1 = 0$

$⇒ 2{\lambda}^{2} - \lambda - 1 = 0$

$⇒ \left(\lambda - 1\right)\left(2\lambda + 1\right) = 0$

$⇒ \lambda = 1, \dfrac{- 1}{2}$

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

{\Delta}_{1} = \left|\begin{cases}5 & 1 & 1 \\ 9 & 2 & 1 \\ \mu & 3 & 1\end{cases}\right|

$⇒ {\Delta}_{1} = \mu - 13$

And for $\lambda = \dfrac{- 1}{2}$ we get,

{\Delta}_{1} = \left|\begin{cases}5 & 1 & 1 \\ 9 & 2 & \dfrac{1}{4} \\ \mu & 3 & \dfrac{- 1}{2}\end{cases}\right| = \dfrac{1}{4}\left|\begin{cases}5 & 1 & 4 \\ 9 & 2 & 1 \\ \mu & 3 & - 2\end{cases}\right| = \dfrac{7}{4}\left(13 - \mu\right)

So, for $\lambda \ne 1 \text{ and } \mu \ne 13$ system has no solution,

Hence, option \left(C\right) is the correct answer.