$\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}\,} \right| = $

The number of distinct real roots of the equaiton, $\left|\begin{array}{*{20}{c}}
{\cos \,\,x}&{\sin \,\,x}&{\sin \,\,x}\\
{\sin \,\,x}&{\cos \,\,x}&{\sin \,\,x}\\
{\sin \,\,x}&{\sin \,\,x}&{\cos \,\,x}
\end{array}\right|\,\, = \,\,0$ in the interval $\left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right]$ is

If $D =$ $\left| {\,\begin{array}{*{20}{c}}{\frac{1}{z}}&{\frac{1}{z}}&{ - \frac{{(x + y)}}{{{z^2}}}}\\{ - \frac{{(y + z)}}{{{x^2}}}}&{\frac{1}{x}}&{\frac{1}{x}}\\{ - \frac{{y(y + z)}}{{{x^2}z}}}&{\frac{{x + 2y + z}}{{xz}}}&{ - \frac{{y(x +y)}}{{x{z^2}}}}\end{array}\,} \right|$ then, the incorrect statement is

$2\,\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2} - bc}&{{b^2} - ac}&{{c^2} - ab}\end{array}\,} \right| = $

By using properties of determinants, show that:

$\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ then $K = $