If $A$ and $B$ are matrices of order $3 \times 3$ and $|A|=5, |B|=3$,then $|3AB|$ is:

The determinant $\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{array} \right|$ is not equal to

If $x, y, z$ are all positive and are the $p$-th, $q$-th, and $r$-th terms of a geometric progression respectively, then the value of the determinant $\left|\begin{array}{lll} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1 \end{array}\right|$ equals:

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

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

By using properties of determinants,show that:
$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$