The value of $\begin{vmatrix} x & x+y & x+y+z \\ 2x & 3x+2y & 4x+3y+2z \\ 3x & 6x+3y & 10x+6y+3z \end{vmatrix}$ is

Given:

$\begin{vmatrix} x & x+y & x+y+z \\ 2x & 3x+2y & 4x+3y+2z \\ 3x & 6x+3y & 10x+6y+3z \end{vmatrix}$

The pattern in this determinant suggests using row operations to create zeros in the first column.

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Perform the row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$:

For Row 2:

First element: $2x - 2(x) = 0$

Second element: $(3x+2y) - 2(x+y)$

$= 3x+2y-2x-2y$

$= x$

Third element: $(4x+3y+2z) - 2(x+y+z)$

$= 4x+3y+2z-2x-2y-2z$

$= 2x+y$

For Row 3:

First element: $3x - 3(x) = 0$

Second element: $(6x+3y) - 3(x+y)$

$= 6x+3y-3x-3y$

$= 3x$

Third element: $(10x+6y+3z) - 3(x+y+z)$

$= 10x+6y+3z-3x-3y-3z$

$= 7x+3y$

The determinant becomes:

$\begin{vmatrix} x & x+y & x+y+z \\ 0 & x & 2x+y \\ 0 & 3x & 7x+3y \end{vmatrix}$

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Expanding along the first column:

$= x \times \begin{vmatrix} x & 2x+y \\ 3x & 7x+3y \end{vmatrix} - 0 + 0$

$= x \times \begin{vmatrix} x & 2x+y \\ 3x & 7x+3y \end{vmatrix}$

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Evaluating the $2 \times 2$ determinant:

$= x \times [x(7x+3y) - (2x+y)(3x)]$

$= x \times [7x^2 + 3xy - 6x^2 - 3xy]$

$= x \times [7x^2 - 6x^2]$

$= x \times x^2$

$= x^3$

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Therefore, the value of the determinant is $x^3$.