The value of the determinant $\left| \begin{array}{ccc} 4 & -6 & 1 \\ -1 & -1 & 1 \\ -4 & 11 & -1 \end{array} \right|$ is

$A$ value of $\theta$ for which the following system of equations has a non-trivial solution is:
$(4 \sin \theta) x - 3y + z = 0$
$x - (6 \cos 2\theta) y + z = 0$
$3x - 12y + 4z = 0$

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:

The value of $\begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$ is

The sum of the distinct values of $x$ for which the matrix $A=\begin{bmatrix} 1 & 1 & x \\ 1 & x & 1 \\ x & 1 & 1 \end{bmatrix}$ has no inverse,is

The value of $a$ for which the system of equations $a^3x + (a + 1)^3y + (a + 2)^3z = 0$,$ax + (a + 1)y + (a + 2)z = 0$,and $x + y + z = 0$ has a non-zero solution is: