$\begin{aligned}& \left|\begin{array}{ccc}1 & 0 & 0 \\0 & \lambda & 0 \\0 & 0 & \lambda^2\end{array}\right|\end{aligned}=\frac{9}{8} \times 81$

$\lambda^3=\frac{9^3}{8} \therefore \lambda=\frac{9}{2}$

$\therefore \frac{\lambda}{3}=\frac{3}{2}$

$\therefore$ Required equation is : $x^2-x\left(\frac{9}{2}+\frac{3}{2}\right) x +\frac{27}{4}=0$ $4 x^2-24 x+27=0$