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

$\left| {\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}} \right| = $

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

If $\left| \begin{matrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{matrix} \right| = (a + b + c)(x + a + b + c)^2$,$x \ne 0$ and $a + b + c \ne 0$,then $x$ is equal to

Show that $\left|\begin{array}{ccc}a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z\end{array}\right|=0$.

If $\left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{{(a + b)}^2}}\end{array}} \right| = k\,abc{(a + b + c)^3}$,then the value of $k$ is