The value of $k$ for which the set of equations $x + ky + 3z = 0, 3x + ky - 2z = 0, 2x + 3y - 4z = 0$ has a non-trivial solution over the set of rationals is

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

If the system of equations
$(k+1)^3 x + (k+2)^3 y = (k+3)^3$
$(k+1) x + (k+2) y = k+3$
$x + y = 1$
is consistent,then the value of $k$ is

$\left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{array} \right| = $

If $f(x) = \left| \begin{array}{ccc} x-3 & 2x^2-18 & 3x^3-81 \\ x-5 & 2x^2-50 & 4x^3-500 \\ 1 & 2 & 3 \end{array} \right|$,then $f(1)f(3) + f(3)f(5) + f(5)f(1)$ is equal to

Let ${a_2},{a_3} \in R$ such that $\left| {{a_2} - {a_3}} \right| = 6$ and $f\left( x \right) = \left| \begin{array}{ccc} 1 & {a_3} & {a_2} \\ 1 & {a_3} & {2{a_2} - x} \\ 1 & {2{a_3} - x} & {a_2} \end{array} \right|, x \in R.$ Then the greatest value of $f(x)$ is