A) 2
B) 4
C) 6
D) 8
Correct Answer: D
Solution :
\[\left| \,\begin{matrix} y+z & x-z & x-y \\ y-z & z+x & y-x \\ z-y & z-x & x+y \\ \end{matrix}\, \right|\]\[=\left| \,\begin{matrix} y+z & x-z & x-y \\ 2y & 2x & 0 \\ 2z & 0 & 2x \\ \end{matrix}\, \right|\] \[{{R}_{2}}\to {{R}_{2}}+{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}+{{R}_{1}}\] \[=4\left| \begin{matrix} y+z & x-z & x-y \\ y & x & 0 \\ z & 0 & x \\ \end{matrix} \right|\] \[=4[(y+z)({{x}^{2}})-(x-z)(xy)\]\[+(x-y)(-zx)]\] \[=4[{{x}^{2}}y+z{{x}^{2}}-{{x}^{2}}y+xyz-z{{x}^{2}}+xyz]\]\[=8xyz\] Hence, \[k=8\].
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