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If $\left|\begin{array}{lll}y+z & x-z & x-y \\ y-z & z-x & y-x \\ z-y & z-x & x+y\end{array}\right|=k x y z$ then the value of $k$ is
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8
$\left|\begin{array}{ccc}y+z & x-z & x-y \\ y-z & z+x & y-x \\ z-y & z-x & x+y\end{array}\right|=\left|\begin{array}{ccc}y+z & x-z & x-y \\ 2 y & 2 x & 0 \\ 2 z & 0 & 2 x\end{array}\right|$ $R_2 \rightarrow R_2+R_1$ and $R_3 \rightarrow R_3+R_1$
$\begin{aligned} & =4\left|\begin{array}{ccc}y+z & x-z & x-y \\ y & x & 0 \\ z & 0 & x\end{array}\right| \\ & =4\left[(y+z)\left(x^2\right)-(x-z)(x y)+(x-y)(-z x)\right] \\ & =4\left[x^2 y+z x^2-x^2 y+x y z-z x^2+x y z\right]=8 x y z\end{aligned}$
Hence, $k=8$
$\begin{aligned} & =4\left|\begin{array}{ccc}y+z & x-z & x-y \\ y & x & 0 \\ z & 0 & x\end{array}\right| \\ & =4\left[(y+z)\left(x^2\right)-(x-z)(x y)+(x-y)(-z x)\right] \\ & =4\left[x^2 y+z x^2-x^2 y+x y z-z x^2+x y z\right]=8 x y z\end{aligned}$
Hence, $k=8$
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