A) \[0,\,\,2a\]
B) \[a,\,\,2a\]
C) \[0,\,\,3a\]
D) none of these
Correct Answer: C
Solution :
Given \[\left| \begin{matrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} 3a-x & a-x & a+x \\ 3a-x & a+x & a-x \\ 3a-x & a-x & a+x \\ \end{matrix} \right|=0\] \[(C\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}})\] \[\Rightarrow \] \[(3a-x)\left| \begin{matrix} 1 & a-x & a-x \\ 1 & a+x & a-x \\ 1 & a-x & a-x \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(3a-x)\left| \begin{matrix} 1 & a-x & a-x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \\ \end{matrix} \right|=0\] \[\left[ \begin{align} & {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\ \end{align} \right]\] \[\Rightarrow \] \[(3a-x)(4{{x}^{2}})=0\] \[\Rightarrow \] \[x=3a,\,\,0\]
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