If $a \ne p, b \ne q, c \ne r$ and $\begin{vmatrix} p & b & c \\ p + a & q + b & 2c \\ a & b & r \end{vmatrix} = 0$,then $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = $

If $\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{matrix} \right| = 5$,then $\left| \begin{matrix} bc^2 - b^2c & a^2c - ac^2 & ab^2 - ba^2 \\ b^2 - c^2 & c^2 - a^2 & a^2 - b^2 \\ c - b & a - c & b - a \end{matrix} \right|$ is equal to

If the determinant $\left| \begin{array}{ccc} a+p & 1+x & u+f \\ b+q & m+y & v+g \\ c+r & n+z & w+h \end{array} \right|$ splits into exactly $K$ determinants of order $3$,each element of which contains only one term,then the value of $K$ is:

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

$\left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} \right| = $

If $\omega$ is one of the imaginary cube roots of unity,then the value of the determinant $\left| \begin{array}{ccc} 1 & \omega^3 & \omega^2 \\ \omega^3 & 1 & \omega \\ \omega^2 & \omega & 1 \end{array} \right|$ is: