If $\left|\begin{array}{l}y+z\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x-z\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x-y\\ y-z\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}z+x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-x\\ z-y\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}z-x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x+y\end{array}\right|=kxyz$ then the value of k is :

Let $A=\left|\begin{array}{ccc}y+z& x-z& x-y\\ y-z& z+x& y-x\\ z-y& z-x& x+y\end{array}\right|$

$\begin{array}{l}\text{ Applying }{\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3\\ =\left|\begin{array}{ccc}2\mathrm{x}& \mathrm{x}-\mathrm{z}& \mathrm{x}-\mathrm{y}\\ 2\mathrm{y}& \mathrm{z}+\mathrm{x}& \mathrm{y}-\mathrm{x}\\ 2\mathrm{z}& \mathrm{z}-\mathrm{x}& \mathrm{x}+\mathrm{y}\end{array}\right|\\ \text{ Applying }{\mathrm{R}}_1\to {\mathrm{R}}_2+{\mathrm{R}}_1,{\mathrm{R}}_3\to {\mathrm{R}}_3+{\mathrm{R}}_1\\ =\left|\begin{array}{ccc}2\mathrm{x}& \mathrm{x}-\mathrm{z}& \mathrm{x}-\mathrm{y}\\ 2(\mathrm{x}+\mathrm{y})& 2\mathrm{x}& 0\\ 2(\mathrm{z}+\mathrm{z})& 0& 2\mathrm{x}\end{array}\right|\end{array}$

On expanding, we get

$\begin{array}{l}=2x\left(4x^2\right)-(x-z)\left[4x\right(x+y\left)\right]+(x-y)[-4x(x+z\left)\right]\\ =8x^2-(x-z)\left(4x^2+4xy\right)-(x-y)\left(4x^2+4xz\right)\\ =8x^3-4x^3-4x^{2}y+4z{x}^2+4xyz-4x^3-4x^{2}z+4y{x}^2+4xyz\\ =8xyz\end{array}$

Given : $A=kxyz\Rightarrow 8xyz=kxyz\Rightarrow k=8$