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 $\left| \begin{array}{ccc} x & 3 & 1 \\ 7 & 6 & z \\ 1 & y & 2 \end{array} \right| + 48$ must be divisible by:

If the determinant of the matrix $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & \beta \\ -b & -\alpha & 0 \end{bmatrix}$ is zero for all $a, b$, then $\alpha + \beta =$

Given that, $a \alpha^2+2 b \alpha+c \neq 0$ and that the system of equations
$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$
has a non-trivial solution, then $a, b$ and $c$ lie in

An equilateral triangle has each of its sides of length $6 \text{ cm}$. If $(x_1, y_1), (x_2, y_2), \text{ and } (x_3, y_3)$ are its vertices,then the value of the determinant $\left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|^2$ is equal to:

Which of the following statements is correct regarding a determinant?

If $\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse,then the real value of $x$ is