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# The repeated factor of $\begin{vmatrix}y+z&x&y\\z+x& z&x\\x+y& y&z\end{vmatrix}$ is

$\begin{array}{1 1}(A)\;x-y&(B)\;y-z\\(C)\;z-x&(D)\;\text{None of these}\end{array}$

Given :
$\begin{vmatrix}y+z&x&y\\z+x& z&x\\x+y& y&z\end{vmatrix}$
Apply $R_1+R_2+R_3=R_1$
$\Delta=\begin{vmatrix}2(x+y+z) &x+y+z&x+y+z\\z+x&z&x\\x+y&y&z\end{vmatrix}$
$\Rightarrow (x+y+z)\begin{vmatrix}2& 1&1\\z+x& z& x\\x+y& y&z\end{vmatrix}$
Apply $C_1=C_1-C_2-C_3$
$\Rightarrow (x+y+z) \begin{vmatrix}0& 1&1\\0&z&x\\x-z&y&z\end{vmatrix}$
$\Rightarrow (x+y+z)(x-z)(x-z)$
$\Rightarrow (x+y+z)(x-z)^2$
Hence (D) is the correct answer.