If $\left| \begin{array}{ccc} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & b+c & -2c \end{array} \right| = \alpha (a+b)(b+c)(c+a) \neq 0$,then $\alpha$ is equal to

Let $a, b, c$ be such that $b + c \ne 0$. If $\left| \begin{array}{ccc} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{array} \right| + \left| \begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} \cdot a & (-1)^{n+1} \cdot b & (-1)^n \cdot c \end{array} \right| = 0$,then $n$ is equal to:

If $a \neq b \neq c$, $\Delta_1=\left|\begin{array}{lll}1 & a^2 & b c \\ 1 & b^2 & c a \\ 1 & c^2 & a b\end{array}\right|$, $\Delta_2=\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|$ and $\frac{\Delta_1}{\Delta_2}=\frac{6}{11}$, then $11(a+b+c)=$

The value of the determinant $\left| \begin{array}{ccc} 2 & 8 & 4 \\ -5 & 6 & -10 \\ 1 & 7 & 2 \end{array} \right|$ is

If $\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$,then $\sin 4\theta$ is equal to:

The determinant $\left| {\begin{array}{*{20}{c}}{1 + a + x}&{a + y}&{a + z}\\{b + x}&{1 + b + y}&{b + z}\\{c + x}&{c + y}&{1 + c + z}\end{array}} \right|$ is equal to: