$\left| {\begin{array}{ccc} 1 & 1+ac & 1+bc \\ 1 & 1+ad & 1+bd \\ 1 & 1+ae & 1+be \end{array}} \right| = $

If $x, y, z \in \mathbb{R}$,then the value of the determinant $\left|\begin{array}{lll}\left(5^{x}+5^{-x}\right)^{2} & \left(5^{x}-5^{-x}\right)^{2} & 1 \\ \left(6^{x}+6^{-x}\right)^{2} & \left(6^{x}-6^{-x}\right)^{2} & 1 \\ \left(7^{x}+7^{-x}\right)^{2} & \left(7^{x}-7^{-x}\right)^{2} & 1\end{array}\right|$ is:

The determinant $\left|\begin{array}{ccc}a^{2}+10 & a b & a c \\ a b & b^{2}+10 & b c \\ a c & b c & c^{2}+10\end{array}\right|$ is

If ${I_1} = \int\limits_1^{\sin \theta } {\frac{x}{{1 + x^2}}} \,dx$ and ${I_2} = \int\limits_1^{\csc \theta } {\frac{{dx}}{{x\left( {{x^2} + 1} \right)}}}$; then the value of $\left| {\begin{array}{*{20}{c}} {{I_1}}&{I_1^2}&{{I_2}} \\ {{e^{{I_1} + {I_2}}}}&{I_2^2}&{ - 1} \\ 1&{I_1^2 + I_2^2}&{ - 1} \end{array}} \right|$ is

By using properties of determinants,show that:
$\left|\begin{array}{ccc}x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4\end{array}\right|=(5x+4)(4-x)^{2}$

$\left| \begin{array}{ccc} 11 & 12 & 13 \\ 12 & 13 & 14 \\ 13 & 14 & 15 \end{array} \right| = $