If $|A|$ denotes the value of the determinant of the square matrix $A$ of order $3$,then $|-2A|=$

If $\Delta _1 = \left| \begin{matrix} b^5c^6(c^3 - b^3) & a^4c^6(a^3 - c^3) & a^4b^5(b^3 - a^3) \\ b^2c^3(b^6 - c^6) & ac^3(c^6 - a^6) & ab^2(a^6 - b^6) \\ b^2c^3(c^3 - b^3) & ac^3(a^3 - c^3) & ab^2(b^3 - a^3) \end{matrix} \right|$ and $\Delta _2 = \left| \begin{matrix} a & b^2 & c^3 \\ a^4 & b^5 & c^6 \\ a^7 & b^8 & c^9 \end{matrix} \right|$,then $\Delta _1 \Delta _2$ is equal to:

If $\theta \in \left(0, \frac{\pi}{2}\right)$,then $\left|\begin{array}{ccc} (\sin \theta+\operatorname{cosec} \theta)^2 & (\sin \theta-\operatorname{cosec} \theta)^2 & 2020 \\ (\cos \theta+\sec \theta)^2 & (\cos \theta-\sec \theta)^2 & 2020 \\ (\tan \theta+\cot \theta)^2 & (\tan \theta-\cot \theta)^2 & 2020 \end{array}\right| = $

Evaluate $\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$

If $A$ is a square matrix of order $3 \times 3$,then $|KA|$ is equal to

By using properties of determinants,show that:
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$