Let $a-2b+c=1$. If $f(x) = \begin{vmatrix} x+a & x+2 & x+1 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3 \end{vmatrix}$,then:

Suppose $D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ and $D' = \begin{vmatrix} a_1 + pb_1 & b_1 + qc_1 & c_1 + ra_1 \\ a_2 + pb_2 & b_2 + qc_2 & c_2 + ra_2 \\ a_3 + pb_3 & b_3 + qc_3 & c_3 + ra_3 \end{vmatrix}$,then

If $\left| \begin{matrix} x - 4 & 2x & 2x \\ 2x & x - 4 & 2x \\ 2x & 2x & x - 4 \end{matrix} \right| = (A + Bx)(x - A)^2$,then the ordered pair $(A, B) = $ . . . . .

Evaluate the determinant: $\left| \begin{array}{ccc} x & 4 & y + z \\ y & 4 & z + x \\ z & 4 & x + y \end{array} \right|$

$A$ 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

The determinant of a skew-symmetric matrix of order $3$ is always: