Let $A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix}$. If $|A|^2 = 25$, then $|\alpha|$ is equal to:

If $a_i^2 + b_i^2 + c_i^2 = 1$ for $(i = 1, 2, 3)$ and $a_i a_j + b_i b_j + c_i c_j = 0$ for $(i \ne j, i, j = 1, 2, 3)$,then the value of $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|^2$ is

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

The number of distinct real roots of $\left| {\begin{array}{*{20}{c}}{\sin x}&{\cos x}&{\cos x}\\{\cos x}&{\sin x}&{\cos x}\\{\cos x}&{\cos x}&{\sin x}\end{array}} \right| = 0$ in the interval $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$ is

The area of a triangle with vertices $(-3, 0)$,$(3, 0)$,and $(0, k)$ is $9$ sq units. Find the value of $k$.

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