Evaluate $\left|\begin{array}{rr}2 & 4 \\ -1 & 2\end{array}\right|$

The cubic $\left| {\begin{array}{*{20}{c}}
  0&{a - x}&{b - x} \\ 
  { - a - x}&0&{c - x} \\ 
  { - b - x}&{ - c - x}&0 
\end{array}} \right| = 0$ has a reperated root in $x$ then,

Three digit numbers $x17, 3y6$ and $12z$ where $x, y, z$ are integers from $0$ to $9$, are divisible by a fixed constant $k$. Then the determinant $\left| {\,\begin{array}{*{20}{c}}x&3&1\\7&6&z\\1&y&2\end{array}\,} \right|$ + $48$ must be divisible by

Find values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$

If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\{2{x^2} + 3x - 1}&{3x}&{3x - 3}\\{{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}\,} \right| = Ax - 12$, then the value of $A $ is

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