If $\left| \begin{matrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{matrix} \right| = (a + b + c)(x + a + b + c)^2$,$x \ne 0$ and $a + b + c \ne 0$,then $x$ is equal to

$A$ 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

Prove that $\Delta = \left| \begin{array}{ccc} a+bx & c+dx & p+qx \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right| = (1-x^2) \left| \begin{array}{ccc} a & c & p \\ b & d & q \\ u & v & w \end{array} \right|$

If $A$ is a square matrix of order $3$,then which of the following statements is true? (where $I$ is the identity matrix)

If $x, y, z$ are all positive and are the $p$-th,$q$-th,and $r$-th terms of a geometric progression respectively,then the value of the determinant $\left|\begin{array}{lll} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1 \end{array}\right|$ equals:

$\left| {\begin{array}{ccc} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{array}} \right| = $