If ${A_1}, {B_1}, {C_1}, \dots$ are respectively the co-factors of the elements ${a_1}, {b_1}, {c_1}, \dots$ of the determinant $\Delta = \begin{vmatrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{vmatrix}$,then $\begin{vmatrix} {B_2} & {C_2} \\ {B_3} & {C_3} \end{vmatrix} = $

Let ${\Delta _1} = \begin{vmatrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{vmatrix}$ and ${\Delta _2} = \begin{vmatrix} {\alpha _1} & {\beta _1} & {\gamma _1} \\ {\alpha _2} & {\beta _2} & {\gamma _2} \\ {\alpha _3} & {\beta _3} & {\gamma _3} \end{vmatrix}$. Then ${\Delta _1} \times {\Delta _2}$ can be expressed as the sum of how many determinants?

Let $\alpha \beta \neq 0$ and $A = \begin{bmatrix} \beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2\alpha \end{bmatrix}$. If $B = \begin{bmatrix} 3\alpha & -9 & 3\alpha \\ -\alpha & 7 & -2\alpha \\ -2\alpha & 5 & -2\beta \end{bmatrix}$ is the matrix of cofactors of the elements of $A$,then $\operatorname{det}(AB)$ is equal to.

For determinant $A = \begin{vmatrix} 1 & 2 & 13 \\ 3 & 0 & 5 \\ 6 & 7 & 11 \end{vmatrix}$,if $p, q, r$ are co-factors of elements $13, 5$ and $11$ respectively,then $p + 3q + 6r = $ . . . . . . .

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

Let $A = \begin{bmatrix} 3 & 1 & 2 \\ 1 & 2 & 0 \\ 1 & 1 & 4 \end{bmatrix}$ and $pC_{11} + 4C_{21} - 5C_{32} = -2$, where $C_{ij}$ denotes the cofactor of an element $a_{ij}$ of matrix $A$. Find the value of $p$.