Show that $\left|\begin{array}{lll}y+z & x & y \\ z+x & z & x \\ x+y & y & z\end{array}\right|=(x+y+z)(x-z)^{2}$
Let $\Delta=\mid y+z \quad x \quad y$
$z+x \quad z \quad x$
$x+y \quad y \quad z \mid$
$\Rightarrow \Delta=\mid 2(x+y+z) \quad x+y+z \quad x+y+z$
$z+x \quad z \quad x$
$\begin{array}{llll}x+y & y & z \mid & \text { [Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \text { ] }\end{array}$
$=(x+y+z) \mid \begin{array}{lll}2 & 1 & 1\end{array}$
$\begin{array}{ccc}z+x & z & x \\ x+y & y & z \mid\end{array}$
$=(\mathrm{x}+y+\mathrm{z}) \mid 0 \quad 1 \quad 1$
$0 \quad \mathrm{z} \quad \mathrm{x}$
$\begin{array}{lll}x-z & \mathrm{y} & \mathrm{z}\end{array}$ [Applying $C_{1} \rightarrow C_{1}-C_{2}-C_{3}$ ]
$=(\mathrm{x}+y+\mathrm{z})\left\{(x-z) \times\left|\begin{array}{ll}1 & 1 \\ z & x\end{array}\right|\right\}$ [Expanding along $C_{1}$ ]
$=(\mathrm{x}+y+\mathrm{z})(x-z)^{2}$