If $x, y$ and $z$ are non-zero distinct numbers, then $\begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}$ is equal to

The determinant is given as:

$\Delta = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}$

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Applying the row operation $R_1 \to R_1 + R_2$:

$\Delta = \begin{vmatrix} x+y+z & x+y+z & x+y+z \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}$

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Taking $(x+y+z)$ as a common factor from the first row:

$\Delta = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}$

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Since the first row $R_1$ and the third row $R_3$ are identical:

$\Delta = (x+y+z) \cdot 0$

$\Delta = 0$

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OR

Given: $\begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}$ where $x$, $y$, and $z$ are non-zero distinct numbers.

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Perform $C_1 \to C_1 - C_2$:

$\begin{vmatrix} (x+y)-(y+z) & y+z & z+x \\ z-x & x & y \\ 1-1 & 1 & 1 \end{vmatrix}$

$= \begin{vmatrix} x-z & y+z & z+x \\ z-x & x & y \\ 0 & 1 & 1 \end{vmatrix}$

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Perform $C_2 \to C_2 - C_3$:

$\begin{vmatrix} x-z & (y+z)-(z+x) & z+x \\ z-x & x-y & y \\ 0 & 1-1 & 1 \end{vmatrix}$

$= \begin{vmatrix} x-z & y-x & z+x \\ z-x & x-y & y \\ 0 & 0 & 1 \end{vmatrix}$

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Expanding along row 3:

Determinant $= 1 \times (-1)^{3+3} \begin{vmatrix} x-z & y-x \\ z-x & x-y \end{vmatrix}$

$= \begin{vmatrix} x-z & y-x \\ z-x & x-y \end{vmatrix}$

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Evaluating the $2 \times 2$ determinant:

$= (x-z)(x-y) - (y-x)(z-x)$

$= (x-z)(x-y) - [-(x-y)(z-x)]$

$= (x-z)(x-y) + (x-y)(z-x)$

$= (x-z)(x-y) + (x-y)[-(x-z)]$

$= (x-z)(x-y) - (x-y)(x-z)$

$= 0$

Therefore, the determinant equals $0$.