question_answer

If \[\left| \begin{matrix}    {{a}^{2}} & {{b}^{2}} & {{c}^{2}}  \\    {{(a+\lambda )}^{2}} & {{(b+\lambda )}^{2}} & {{(c+\lambda )}^{2}}  \\    {{(a-\lambda )}^{2}} & {{(b-\lambda )}^{2}} & {{(c-\lambda )}^{2}}  \\ \end{matrix} \right|=k\lambda \left| \begin{matrix}    {{a}^{2}} & {{b}^{2}} & {{c}^{2}}  \\    a & b & c  \\    1 & 1 & 1  \\ \end{matrix} \right|\],\[\lambda \ne 0\], then k is equal to

A)  \[-4\lambda \] abc                     

B)  \[4{{\lambda }^{2}}\]

C)  \[-4{{\lambda }^{2}}\]                                   

D)  \[4\lambda \] abc