If ${A_1},{B_1},{C_1}$…. are respectively the co-factors of the elements ${a_1},{b_1},{c_1}$,…… of the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$, then $\left| {\begin{array}{*{20}{c}}{{B_2}}&{{C_2}}\\{{B_3}}&{{C_3}}\end{array}} \right| = $

The sum of the products of the elements of any row of a determinant $A$ with the same row is always equal to

If ${\Delta _1} = \left| {\begin{array}{*{20}{c}}
  {{b^5}{c^6}\left( {{c^3} – {b^3}} \right)}&{{a^4}{c^6}\left( {{a^3} – {c^3}} \right)}&{{a^4}{b^5}\left( {{b^3} – {a^3}} \right)} \\ 
  {{b^2}{c^3}\left( {{b^6} – {c^6}} \right)}&{a{c^3}\left( {{c^6} – {a^6}} \right)}&{a{b^2}\left( {{a^6} – {b^6}} \right)} \\ 
  {{b^2}{c^3}\left( {{c^3} – {b^3}} \right)}&{a{c^3}\left( {{a^3} – {c^3}} \right)}&{a{b^2}\left( {{b^3} – {a^3}} \right)} 
\end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}}
  a&{{b^2}}&{{c^3}} \\ 
  {{a^4}}&{{b^5}}&{{c^6}} \\ 
  {{a^7}}&{{b^8}}&{{c^9}} 
\end{array}} \right|$ then ${\Delta _1}{\Delta _2}$ is equal to

If value of a third order determinant is $11$, then the value of the square of the determinant formed by the cofactors will be

If $a_i^2 + b_i^2 + c_i^2 = 1,\,\,(i = 1,2,3)$ and ${a_i}{a_j} + {b_i}{b_j} + {c_i}{c_j} = 0$ $(i \ne j,i,j = 1,2,3)$ then the value of ${\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}\,} \right|^2}$ is