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

For all values of $A,B,C$ and $P,Q,R$, the value of $\left| {\,\begin{array}{*{20}{c}}{\cos (A - P)}&{\cos (A - Q)}&{\cos (A - R)}\\{\cos (B - P)}&{\cos (B - Q)}&{\cos (B - R)}\\{\cos (C - P)}&{\cos (C - Q)}&{\cos (C - R)}\end{array}\,} \right|$ is

The system of linear equation $x + y + z = 2, 2x + 3y + 2z = 5$, $2x + 3y + (a^2 -1)\,z = a + 1$ then

$\left| {\,\begin{array}{*{20}{c}}{1/a}&{{a^2}}&{bc}\\{1/b}&{{b^2}}&{ca}\\{1/c}&{{c^2}}&{ab}\end{array}\,} \right| = $

Find area of the triangle with vertices at the point given in each of the following: $(-2,-3),(3,2),(-1,-8)$

If $\left| \begin{array}{*{20}{c}}
{ - 2a}&{a + b}&{a + c}\\
{b + a}&{ - 2b}&{b + c}\\
{c + a}&{b + c}&{ - 2c}
\end{array}\right|$ $ = \alpha \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) \ne 0$ then $\alpha $ is equal to