The value of […] is

The correct answer is Option (4) → $x^3$ **

Determinant:

$\displaystyle D=\begin{vmatrix} x & x+y & x+y+z\\[4pt] 2x & 3x+2y & 4x+3y+2z\\[4pt] 3x & 6x+3y & 10x+6y+3z \end{vmatrix}$

Row operations: $R_{2}\to R_{2}-2R_{1},\ R_{3}\to R_{3}-3R_{1}$

After operations the matrix becomes

$\displaystyle \begin{vmatrix} x & x+y & x+y+z\\[4pt] 0 & x & 2x+y\\[4pt] 0 & 3x & 7x+3y \end{vmatrix}$

Expand along first column:

$\displaystyle D = x\begin{vmatrix} x & 2x+y\\[4pt] 3x & 7x+3y\end{vmatrix}$

$\displaystyle = x\bigl(x(7x+3y)-(2x+y)(3x)\bigr)$

$\displaystyle = x\bigl(7x^{2}+3xy-6x^{2}-3xy\bigr)=x\cdot x^{2}=x^{3}$