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(D) Let $A = \begin{bmatrix} -x & 14x & 7x \ 0 & 1 & 0 \ x & -4x & -2x \end{bmatrix}$. The inverse $A^{-1}$ is given as $\begin{bmatrix} 2 & 0 & 7 \

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(D) Let $A = \begin{bmatrix} -x & 14x & 7x \ 0 & 1 & 0 \ x & -4x & -2x \end{bmatrix}$. The inverse $A^{-1}$ is given as $\begin{bmatrix} 2 & 0 & 7 \ 0 & 1 & 0 \ 1 & -2 & 1 \end{bmatrix}$.We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix.Multiplying the first row of $A$ by the first column of $A^{-1}$:$(-x)(2) + (14x)(0) + (7x)(1) = 1$ (since the element at $(1,1)$ of $I$ is $1$)$-2x + 7x = 1 \implies 5x = 1 \implies x = 1/5$.Now,we need to evaluate the determinant $D = \left|\begin{array}{ccc} x & x+1 & x+2 \ x+1 & x+2 & x+3 \ x+2 & x+3 & x+4 \end{array}ight|$.Apply row operations: $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$.$D = \left|\begin{array}{ccc} x & x+1 & x+2 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{array}ight|$.Since two rows are identical,the determinant is $0$.

If the inverse of $\begin{bmatrix} -x & 14x & 7x \\ 0 & 1 & 0 \\ x & -4x & -2x \end{bmatrix}$ is $\begin{bmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$,then $\left|\begin{array}{ccc} x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \end{array}\right| = $

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