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(D) Let $\Delta = \left|\begin{array}{ccc}x+y & y+z & z+x \ z & x & y \ 1 & 1 & 1\end{array}ight|$.Applying the column operation $C_1 	o C_1 + C_

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(D) Let $\Delta = \left|\begin{array}{ccc}x+y & y+z & z+x \ z & x & y \ 1 & 1 & 1\end{array}ight|$.Applying the column operation $C_1 	o C_1 + C_2$:$\Delta = \left|\begin{array}{ccc}x+y+z & y+z & z+x \ z+x & x & y \ 2 & 1 & 1\end{array}ight|$.This does not immediately simplify to zero. Let us apply $C_1 	o C_1 + C_2 + C_3$:$\Delta = \left|\begin{array}{ccc}2(x+y+z) & y+z & z+x \ x+y+z & x & y \ 3 & 1 & 1\end{array}ight|$.Alternatively,apply $C_1 	o C_1 + C_2$ and then observe the rows. Let us perform $C_1 	o C_1 + C_2$:$\Delta = \left|\begin{array}{ccc}x+y+z & y+z & z+x \ x+z & x & y \ 2 & 1 & 1\end{array}ight|$.Actually,applying $C_1 	o C_1 + C_2$ is not helpful. Let us apply $C_1 	o C_1 + C_2 + C_3$ is also not helpful. Let us apply $R_1 	o R_1 + R_2$:$\Delta = \left|\begin{array}{ccc}x+y+z & x+y+z & x+y+z \ z & x & y \ 1 & 1 & 1\end{array}ight|$.Now,take $(x+y+z)$ as a common factor from $R_1$:$\Delta = (x+y+z) \left|\begin{array}{ccc}1 & 1 & 1 \ z & x & y \ 1 & 1 & 1\end{array}ight|$.Since $R_1$ and $R_3$ are identical,the value of the determinant is $0$.

$\left|\begin{array}{ccc}x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1\end{array}\right|=$ . . . . . . .

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