A) \[2xyz\]
B) 1
C) \[xyz\]
D) \[{{x}^{2}}{{y}^{2}}{{z}^{2}}\]
Correct Answer: B
Solution :
\[\left| \,\begin{matrix} y+z & x & y \\ z+x & z & x \\ x+y & y & z \\ \end{matrix}\, \right|\]= \[(x+y+z)\,\left| \,\begin{matrix} 2 & 1 & 1 \\ z+x & z & x \\ x+y & y & z \\ \end{matrix}\, \right|\] by \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\] = \[(x+y+z)\,\left| \,\begin{matrix} 1 & 1 & 1 \\ x & z & x \\ x & y & z \\ \end{matrix}\, \right|\] ; by \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}}\] = \[(x+y+z)\,.\,\{({{z}^{2}}-xy)-(xz-{{x}^{2}})+(xy-xz)\}\] = \[(x+y+z)\,{{(x-z)}^{2}}\] \[\Rightarrow \] \[k=1\]. Trick: Put\[x=1,\,y=2\], \[z=3\], then \[\left| \,\begin{matrix} 5 & 1 & 2 \\ 4 & 3 & 1 \\ 3 & 2 & 3 \\ \end{matrix}\, \right|\,=\,5(7)\,-1(12-3)+2(8-9)\] = \[35-9-2=24\]& \[(x+y+z)\,{{(x-z)}^{2}}=(6)\,{{(-2)}^{2}}=24\] \[\therefore \,\,k=\frac{24}{24}=1\].
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