If $a, b, c$ are real then the value of determinant $\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac}\\{ab}&{{b^2} + 1}&{bc}\\{ac}&{bc}&{{c^2} + 1}\end{array}}\right|$ $= 1$ if

If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 3}\\{x + 2}&{x + 3}&{x + 4}\\{x + a}&{x + b}&{x + c}\end{array}\,} \right| = 0$, then $a,b,c$ are in

Value of $\left| {\begin{array}{*{20}{c}}
  {{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\ 
  {{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\ 
  {{c^2}}&{{c^2}}&{{{(a + b)}^2}} 
\end{array}} \right|$ is equal to

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{b + c}&{c + a}&{a + b}\\{b + c - a}&{c + a - b}&{a + b - c}\end{array}\,} \right|$ is

Prove that

$\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$

If $f(x) = \left| {\begin{array}{*{20}{c}}1&x&{x + 1}\\{2x}&{x(x - 1)}&{(x + 1)x}\\{3x(x - 1)}&{x(x - 1)(x - 2)}&{(x + 1)x(x - 1)}\end{array}} \right|$ then $f(100)$ is equal to