If $a, b, c$ are all different and $\left| \begin{array}{ccc} a & a^3 & a^4 - 1 \\ b & b^3 & b^4 - 1 \\ c & c^3 & c^4 - 1 \end{array} \right| = 0$,then the value of $abc(ab + bc + ca)$ is

Verify Property $1$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$

Evaluate the determinant: $\left| {\begin{array}{*{20}{c}}{{b^2} - ab}&{b - c}&{bc - ac}\\{ab - {a^2}}&{a - b}&{{b^2} - ab}\\{bc - ac}&{c - a}&{ab - {a^2}}\end{array}} \right|$

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

$A$ value of $\theta$ in $\left(0, \frac{\pi}{2}\right)$ satisfying $\left|\begin{array}{ccc}1+\sin ^2 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta\end{array}\right|=0$ is

Let $A = \begin{bmatrix} 1 + x^2 - y^2 - z^2 & 2(xy + z) & 2(zx - y) \\ 2(xy - z) & 1 + y^2 - z^2 - x^2 & 2(yz + x) \\ 2(zx + y) & 2(yz - x) & 1 + z^2 - x^2 - y^2 \end{bmatrix}$. Then $\det(A)$ is equal to: