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

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

If $A = \left[ {\begin{array}{*{20}{c}}
1&{\sin \,\theta }&1\\
{ - \,\sin \,\theta }&1&{\sin \,\theta }\\
{ - 1}&{ - \,\sin \,\theta }&1
\end{array}} \right];$ then for all $\theta \, \in \,\left( {\frac{{3\pi }}{4},\frac{{5\pi }}{4}} \right),$ det $(A)$ lies in the interval

If a system of the equation ${(\alpha + 1)^3}x + {(\alpha + 2)^3}y - {(\alpha + 3)^3} = 0$ and $(\alpha + 1)x + (\alpha + 2)y - (\alpha + 3) = 0,x + y - 1 = 0$ is constant. what is the value of $\alpha $

Let $A=\left(\begin{array}{ccc}{[x+1]} & {[x+2]} & {[x+3]} \\ {[x]} & {[x+3]} & {[x+3]} \\ {[x]} & {[x+2]} & {[x+4]}\end{array}\right),$ where $[t]$ denotes the greatest integer less than or equal to $\mathrm{t}$. If $\operatorname{det}(\mathrm{A})=192$, then the set of values of $\mathrm{x}$ is the interval

If the system of equations $x+y+z=6 \,; \,2 x+5 y+\alpha z=\beta  \,; \, x+2 y+3 z=14$ has infinitely many solutions, then $\alpha+\beta$ is equal to.