If | cases x + 1 & x & x x & x + λ & x x & x & x + λ ^ 2 cases | =...

Given that, \left|\begin{cases}x + 1 & x & x \\ x & x + \lambda & x \\ x & x & x + {\lambda}^{2}\end{cases}\right| = \dfrac{9}{8}\left(103x + 81\right)

Put $x = 0$ as $x \in R$

⇒ \left|\begin{cases}1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & {\lambda}^{2}\end{cases}\right| = \dfrac{9}{8}\left(103\left(0\right) + 81\right)

On expanding the determinant we get,

$⇒ {\lambda}^{3} = \dfrac{9}{8} \times 81$

$⇒ {\lambda}^{3} = \dfrac{{9}^{3}}{{2}^{3}}$

$⇒ \lambda = \dfrac{9}{2}$ and $\dfrac{\lambda}{3} = \dfrac{9}{6}$

Now the required quadratic equation is ${x}^{2} - \left(\dfrac{9}{2} + \dfrac{9}{6}\right)x + \dfrac{9}{2} \times \dfrac{9}{6} = 0$

$⇒ {x}^{2} - \left(6\right)x + \dfrac{27}{4} = 0$

$⇒ 4{x}^{2} - 24x + 27 = 0$

Hence, the required quadratic equation is $4{x}^{2} - 24x + 27 = 0$