A) \[x=0,x=4a\]
B) \[x=0,x=a\]
C) \[x=0,\text{ }x=2a\]
D) \[x=0,\text{ }x=3a\]
Correct Answer: D
Solution :
\[\left| \begin{matrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} 3a-x & a-x & a-x \\ 3a-x & a+x & a-x \\ 3a-x & a-x & a+x \\ \end{matrix} \right|=0\] \[[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}]\] \[\Rightarrow \] \[(3a-x)\left| \begin{matrix} 1 & a-x & a-x \\ 1 & a+x & a-x \\ 1 & a-x & a+x \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(3a-x)\left| \begin{matrix} 1 & a-x & a-x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \\ \end{matrix} \right|=0\] \[\left[ \begin{align} & {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\ \end{align} \right]\] \[\Rightarrow \] \[(3a-x)[1{{(4x)}^{2}}]=0\] \[\Rightarrow \] \[4{{x}^{2}}(3a-x)=0\] \[\Rightarrow \] \[x=0,3a\]
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