If $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right| = 10 k x y z \left(3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$,then $k = \text{ . . . . . . }$ (where $x, y, z \neq 0$ and $3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \neq 0$).

If $\left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0$,then the area of the triangle whose vertices are $\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right), \left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right), \left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right)$ is:

The greatest value of $c \in R$ for which the system of linear equations $x - cy - cz = 0$,$cx - y + cz = 0$,$cx + cy - z = 0$ has a non-trivial solution,is

The matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is not invertible only if $a =$

If $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ are the vertices of a triangle whose area is $k$ square units,then $\left|\begin{array}{ccc}x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4\end{array}\right|^{2}$ is (in $k^{2}$)

If $f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}\right|$, then $f(2012)$ is equal to: