Evaluate the determinant: $\left| \begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array} \right|$

If $a, b$ and $c$ are real numbers such that $a^2+b^2+c^2-ab-bc-ac \leq 0$,then the value of the determinant $\left|\begin{array}{ccc} (a-b+1)^5 & b^7-c^7 & c^9-a^9 \\ a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \\ a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3)^1 \end{array}\right|$ is:

Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$.

Let $0 \neq a \in \mathbb{Z}$ and $A = \begin{bmatrix} a & a & a-y \\ a & a+x & a \\ a & a & a \end{bmatrix}$ be a matrix. Then,the equation $\det(A) = 16$ represents:

If $a, b, c$ are positive and not all equal,then the value of the determinant $\left| \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array} \right|$ is

Let $A=\begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}$. If $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(3A)))=2^m \cdot 3^n$,where $m, n \in N$,then $m+n$ is equal to: