The value of the determinant

Given: $\Delta=\left|\begin{array}{ccc}0 & x-y & y-z \\ y-x & 0 & z-x \\ z-y & x-z & 0\end{array}\right|$

Let $A=\left[\begin{array}{ccc}0 & x-y & y-z \\ y-x & 0 & z-x \\ z-y & x-z & 0\end{array}\right]$

$A^{T}=\left[\begin{array}{ccc}0 & x-y & y-z \\ y-x & 0 & z-x \\ z-y & x-z & 0\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}0 & y-x & z-y \\ x-y & 0 & x-z \\ y-z & z-x & 0\end{array}\right]$

$=(-1)\left[\begin{array}{ccc}0 & x-y & y-z \\ y-x & 0 & z-x \\ z-y & x-z & 0\end{array}\right]$

$=-A$

Since, $A^{T}=-A$

Therefore, $A$ is a skew symmetric matrix.

Thus, $|A|=0$

$\Rightarrow\left|\begin{array}{ccc}0 & x-y & y-z \\ y-x & 0 & z-x \\ z-y & x-z & 0\end{array}\right|=0$

$\Rightarrow \Delta=0$

Hence, the value of the determinant $\Delta=\left|\begin{array}{ccc}0 & x-y & y-z \\ y-x & 0 & z-x \\ z-y & x-z & 0\end{array}\right|$ is $\underline{0}$.