$\left| {\,\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\{a + 2b}&{a + 3b}&{a + 4b}\\{a + 4b}&{a + 5b}&{a + 6b}\end{array}\,} \right| = $

If $a, b, c,$ are non zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{ab}&{ac}\\
{ab}&{{c^2} + {a^2}}&{bc}\\
{ac}&{bc}&{{a^2} + {b^2}}
\end{array}} \right| = k{a^2}{b^2}{c^2},$ then $k$ is equal to

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

By using properties of determinants, show that:

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

Without expanding the determinant, prove that

$\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c\end{array}\right|=a^{3}$