If $a, b, c$ are three complex numbers such that $a^2 + b^2 + c^2 = 0$ and  $\left| {\begin{array}{*{20}{c}}
{\left( {{b^2} + {c^2}} \right)}&{ab}&{ac}\\
{ab}&{\left( {{c^2} + {a^2}} \right)}&{bc}\\
{ac}&{bc}&{\left( {{a^2} + {b^2}} \right)}
\end{array}} \right| = K{a^2}{b^2}{c^2}$ then value of $K$ is

For all values of $A,B,C$ and $P,Q,R$, the value of $\left| {\,\begin{array}{*{20}{c}}{\cos (A – P)}&{\cos (A – Q)}&{\cos (A – R)}\\{\cos (B – P)}&{\cos (B – Q)}&{\cos (B – R)}\\{\cos (C – P)}&{\cos (C – Q)}&{\cos (C – R)}\end{array}\,} \right|$ is

If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$,then $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :

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

If $\left| {\begin{array}{*{20}{c}}
  {^9{C_4}}&{^9{C_5}}&{^{10}{C_r}} \\ 
  {^{10}{C_6}}&{^{10}{C_7}}&{^{11}{C_{r + 2}}} \\ 
  {^{11}{C_8}}&{^{11}{C_9}}&{^{12}{C_{r + 4}}} 
\end{array}} \right| = 0$ then $r$ is equal to