The value of $\left| {\,\begin{array}{*{20}{c}}{265}&{240}&{219}\\{240}&{225}&{198}\\{219}&{198}&{181}\end{array}\,} \right|$ is equal to

If $ab + bc + ca = 0$ and $\left| {\,\begin{array}{*{20}{c}}{a – x}&c&b\\c&{b – x}&a\\b&a&{c – x}\end{array}\,} \right| = 0$, then one of the value of $x$ is

By using properties of determinants, show that:

$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

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

Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If

$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$

then value of $\lambda^{2}$ is equal to $…..$