A value of $\theta  \in  (0, \pi /3)$, for which $\left| {\begin{array}{*{20}{c}}
  {1 + {{\cos }^2}\,\theta }&{{{\sin }^2}\,\theta }&{4\,\cos \,6\theta } \\ 
  {{{\cos }^2}\,\theta }&{1 + {{\sin }^2}\,\theta }&{4\,\cos \,6\theta } \\ 
  {{{\cos }^2}\,\theta }&{{{\sin }^2}\,\theta }&{1 + 4\,\cos \,6\theta } 
\end{array}} \right| = 0$, is

Without expanding the determinant, prove that

$\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

$\left| {\,\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\{a + 2b}&{a + 3b}&{a + 4b}\\{a + 4b}&{a + 5b}&{a + 6b}\end{array}\,} \right| = $

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 $a,b,c$ are unequal what is the condition that the value of the following determinant is zero $\Delta = \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} + 1}\\b&{{b^2}}&{{b^3} + 1}\\c&{{c^2}}&{{c^3} + 1}\end{array}\,} \right|$