$\left| {\begin{array}{*{20}{c}}0&a&{ – b}\\{ – a}&0&c\\b&{ – c}&0\end{array}} \right| = $

Three digit numbers $x17, 3y6$ and $12z$ where $x, y, z$ are integers from $0$ to $9$, are divisible by a fixed constant $k$. Then the determinant $\left| {\,\begin{array}{*{20}{c}}x&3&1\\7&6&z\\1&y&2\end{array}\,} \right|$ + $48$ must be divisible by

If $\alpha ,\beta \ne 0$ and $f\left( n \right) = {\alpha ^n} + {\beta ^n}$ and $\left| {\begin{array}{*{20}{c}}3&{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}\\{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}\\{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}&{1 + f\left( 4 \right)}\end{array}} \right|\; = K{\left( {1 – \alpha } \right)^2}$ ${\left( {1 – \beta } \right)^2}{\left( {\alpha – \beta } \right)^2}$ ,then $K=$ . . . . . .

Given the system of equation $a(x + y + z)=x,b(x + y + z) = y, c(x + y + z) = z$ where $a,b,c$  are non-zero real numbers. If the real numbers $x,y,z$ are such that $xyz \neq 0,$ then  $(a + b + c)$ is equal to-

Let $\theta \in\left(0, \frac{\pi}{2}\right)$. If the system of linear equations

$\left(1+\cos ^{2} \theta\right) x+\sin ^{2} \theta y+4 \sin 3 \theta z=0$

$\cos ^{2} \theta x+\left(1+\sin ^{2} \theta\right) y+4 \sin 3 \theta z=0$

$\cos ^{2} \theta x+\sin ^{2} \theta y+(1+4 \sin 3 \theta) z=0$

has a non-trivial solution, then the value of $\theta$ is :