If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 3}\\{x + 2}&{x + 3}&{x + 4}\\{x + a}&{x + b}&{x + c}\end{array}\,} \right| = 0$, then $a,b,c$ are in

The determinant $\left| {\begin{array}{*{20}{c}}{1\, + \,a\, + \,x}&{a\, + \,y}&{a\, + \,z}\\{b\, + \,x}&{1\, + \,b\, + \,y}&{b\, + \,z}\\{c\, + \,x}&{c\, + \,y}&{1\, + \,c\, + \,z}\end{array}} \right|$ $=$

If $a + b + c = 0$, then the solution of the equation $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$ is

Let $A$ be a $3 \times 3$ matrix with $\operatorname{det}( A )=4$. Let $R _{ i }$ denote the $i ^{\text {th }}$ row of $A$. If a matrix $B$ is obtained by performing the operation $R _{2} \rightarrow 2 R _{2}+5 R _{3}$ on $2 A ,$ then $\operatorname{det}( B )$ is equal to ...... .

$\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + y}&1\\1&1&{1 + z}\end{array}\,} \right| = $

Show that

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