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

Area of the triangle formed by points $(102, -4)$,$(105, -2)$,and $(103, -3)$ is:

If $\begin{bmatrix} 2 + x & 3 & 4 \\ 1 & -1 & 2 \\ x & 1 & -5 \end{bmatrix}$ is a singular matrix,then $x$ is

If $\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|$,then the two triangles with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(a_1, b_1), (a_2, b_2), (a_3, b_3)$ must be:

For all values of $A, B, C$ and $P, Q, R$,the value of $\left| \begin{array}{ccc} \cos(A-P) & \cos(A-Q) & \cos(A-R) \\ \cos(B-P) & \cos(B-Q) & \cos(B-R) \\ \cos(C-P) & \cos(C-Q) & \cos(C-R) \end{array} \right|$ is

The value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \end{array} \right|$ is