If $$\begin{vmatrix} x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^2 \end{vmatrix} = \frac{9}{8}(103x + 81)$$, then $$\lambda$$, $$\frac{\lambda}{3}$$ are the roots of the equation

We need to find $$\lambda$$ such that the given determinant equals $$\frac{9}{8}(103x + 81)$$.

$$ D = \begin{vmatrix} x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^2 \end{vmatrix} $$

Apply $$R_1 \to R_1 - R_2$$ and $$R_2 \to R_2 - R_3$$:

$$R_1 - R_2 = (1,\; -\lambda,\; 0)$$

$$R_2 - R_3 = (0,\; \lambda,\; -\lambda^2)$$

$$R_3 = (x,\; x,\; x+\lambda^2)$$

$$ D = \begin{vmatrix} 1 & -\lambda & 0 \\ 0 & \lambda & -\lambda^2 \\ x & x & x+\lambda^2 \end{vmatrix} $$

$$ D = 1 \cdot \begin{vmatrix} \lambda & -\lambda^2 \\ x & x+\lambda^2 \end{vmatrix} + x \cdot \begin{vmatrix} -\lambda & 0 \\ \lambda & -\lambda^2 \end{vmatrix} $$

First minor: $$\lambda(x + \lambda^2) - (-\lambda^2)(x) = \lambda x + \lambda^3 + \lambda^2 x$$

Second minor: $$(-\lambda)(-\lambda^2) - (0)(\lambda) = \lambda^3$$

$$ D = \lambda x + \lambda^3 + \lambda^2 x + x\lambda^3 $$

$$ = x\lambda(1 + \lambda + \lambda^2) + \lambda^3 $$

When $$x = 0$$: $$D = \begin{vmatrix} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda^2 \end{vmatrix} = \lambda^3$$. Our formula gives $$\lambda^3$$ $$\checkmark$$

$$ x\lambda(1 + \lambda + \lambda^2) + \lambda^3 = \frac{927x}{8} + \frac{729}{8} $$

Comparing constant terms: $$\lambda^3 = \frac{729}{8} = \left(\frac{9}{2}\right)^3$$, so $$\lambda = \frac{9}{2}$$

Verification of coefficient of $$x$$: $$\lambda(1 + \lambda + \lambda^2) = \frac{9}{2}\left(1 + \frac{9}{2} + \frac{81}{4}\right) = \frac{9}{2} \times \frac{103}{4} = \frac{927}{8}$$ $$\checkmark$$

Roots: $$\frac{9}{2}$$ and $$\frac{3}{2}$$

Sum of roots: $$\frac{9}{2} + \frac{3}{2} = 6$$

Product of roots: $$\frac{9}{2} \times \frac{3}{2} = \frac{27}{4}$$

Equation: $$x^2 - 6x + \frac{27}{4} = 0$$, i.e., $$4x^2 - 24x + 27 = 0$$

The correct answer is Option D: $$4x^2 - 24x + 27 = 0$$.