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

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$

If ${a_1},{a_2},{a_3},........,{a_n},......$ are in G.P. and ${a_i} > 0$  for each $i$, then the value of the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 2}}}&{\log {a_{n + 4}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 8}}}&{\log {a_{n + 10}}}\\{\log {a_{n + 12}}}&{\log {a_{n + 14}}}&{\log {a_{n + 16}}}\end{array}} \right|$ is equal to

If $ab + bc + ca = 0$ and $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$, then one of the value of $x$ is

For a non - zero, real $a, b$ and $c$ $\left| {\begin{array}{*{20}{c}}{\frac{{{a^2} + {b^2}}}{c}}&c&c\\a&{\frac{{{b^2} + {c^2}}}{a}}&a\\b&b&{\frac{{{c^2} + {a^2}}}{b}} \end{array}} \right|$ $= \alpha \, abc$, then the values of $\alpha$ is

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