If $\left| {\begin{array}{*{20}{c}}
  {{a^2}}&{{b^2}}&{{c^2}} \\ 
  {{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\ 
  {{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}} 
\end{array}} \right|$ $ = \,k\lambda \,\,\left| {{\mkern 1mu} {\mkern 1mu} \begin{array}{*{20}{c}}
  {{a^2}}&{{b^2}}&{{c^2}} \\
  a&b&c \\
  1&1&1
\end{array}} \right|,\lambda \, \ne \,0$ then $k$ is equal to

If $a+x=b+y=c+z+1,$ where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to

If $\left| {\begin{array}{*{20}{c}}   {a - b}&{b - c}&{c - a} \\    {b - c}&{c - a}&{a - b} \\    {c - a + 1}&{a - b}&{b - c}  \end{array}} \right| = 0$ ,$\left( {a,b,c \in R - \left\{ 0 \right\}} \right),$ then

The value of $\left| {\,\begin{array}{*{20}{c}}{265}&{240}&{219}\\{240}&{225}&{198}\\{219}&{198}&{181}\end{array}\,} \right|$ is equal to

If the determinant $\left| {\begin{array}{*{20}{c}}{a\, + \,p}&{1\, + \,x}&{u\, + \,f}\\ {b\, + \,q}&{m\, + \,y}&{v\, + \,g}\\{c\, + \,r}&{n\, + \,z}&{w\, + \,h}\end{array}} \right|$ splits into exactly $K$ determinants of order $3$, each element of which contains only one term, then the value of $K$, is

Evaluate $\Delta=\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$