1. Interpret the determinant equation

We have

$$\begin{vmatrix} x+1 & x & x\\ x & x+\lambda & x\\ x & x & x+\lambda^2 \end{vmatrix} = \frac{9}{8}(103x+81).$$

We first simplify the determinant.

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2. Write the matrix in a convenient form

Observe that the matrix is

$$\begin{pmatrix} x+1 & x & x\\ x & x+\lambda & x\\ x & x & x+\lambda^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \lambda^2 \end{pmatrix} + \begin{pmatrix} x & x & x\\ x & x & x\\ x & x & x \end{pmatrix}.$$

So this is of the form $D + xJ,$ where $D=\operatorname{diag}(1,\lambda,\lambda^2), \qquad J=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}.$

But it is easier here to expand directly.

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3. Compute the determinant

Let

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

Apply row operations: $R_2 \to R_2-R_1, \qquad R_3 \to R_3-R_1.$ Then

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

Now expand along the first row:

$$\Delta=(x+1) \begin{vmatrix} \lambda & 0\\ 0 & \lambda^2 \end{vmatrix} -x \begin{vmatrix} -1 & 0\\ -1 & \lambda^2 \end{vmatrix} +x \begin{vmatrix} -1 & \lambda\\ -1 & 0 \end{vmatrix}.$$

Compute each minor:

$$\begin{vmatrix} \lambda & 0\\ 0 & \lambda^2 \end{vmatrix}=\lambda^3,$$ $$\begin{vmatrix} -1 & 0\\ -1 & \lambda^2 \end{vmatrix}=-\lambda^2,$$ $$\begin{vmatrix} -1 & \lambda\\ -1 & 0 \end{vmatrix}=\lambda.$$

Hence

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

So

$$\Delta=x(\lambda^3+\lambda^2+\lambda)+\lambda^3.$$

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4. Compare coefficients with the given RHS

Given

$$\Delta=\frac{9}{8}(103x+81)=\frac{927}{8}x+\frac{729}{8}.$$

Therefore,

$$\lambda^3=\frac{729}{8}.$$

So

$$\lambda=\frac{9}{2}.$$

Also,

$$\lambda^3+\lambda^2+\lambda=\frac{927}{8}.$$

Check with $\lambda=\frac92$:

$$\lambda^3+\lambda^2+\lambda =\frac{729}{8}+\frac{81}{4}+\frac92 =\frac{729+162+36}{8} =\frac{927}{8},$$

which matches.

So the two numbers are

$$\lambda=\frac92, \qquad \frac{\lambda}{3}=\frac32.$$

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5. Form the quadratic whose roots are $\lambda$ and $\lambda/3$

Roots are

$$\frac92, \frac32.$$

Their sum is

$$\frac92+\frac32=6,$$

and product is

$$\frac92\cdot\frac32=\frac{27}{4}.$$

Hence the quadratic is

$$x^2-6x+\frac{27}{4}=0.$$

Multiplying by 4,

$$4x^2-24x+27=0.$$

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6. Match with the options

This corresponds to:

Option B: $4x^2-24x+27=0.$

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7. Compare with stored correct answer

Stored correct answer is B, which matches our derived answer.