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

Show that

$\Delta=\left|\begin{array}{ccc}
(y+z)^{2} & x y & z x \\
x y & (x+z)^{2} & y z \\
x z & y z & (x+y)^{2}
\end{array}\right|=2 x y z(x+y+z)^{3}$

By using properties of determinants, show that:

$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

If $a, b $ and $ c$ are non zero numbers, then $\Delta = \left| {\,\begin{array}{*{20}{c}}{{b^2}{c^2}}&{bc}&{b + c}\\{{c^2}{a^2}}&{ca}&{c + a}\\{{a^2}{b^2}}&{ab}&{a + b}\end{array}\,} \right|$ is equal to

Let $a-2 b+c=1$

If $f(x)=\left|\begin{array}{lll}{x+a} & {x+2} & {x+1} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|,$ then

If the system of equations $ax + y + z = 0$, $x + by + z = 0$ and $x + y + cz = 0 $, where $a,b,c \ne 1,$ has a non trivial solution, then the value of $\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}}$is