Number of triplets of $a, b \, \& \,c$ for which the system of equations,$ax - by = 2a - b$ and $(c + 1) x + cy = 10 - a + 3 b$ has infinitely many solutions and $x = 1, y = 3$ is one of the solutions, is :

$\left| {\,\begin{array}{*{20}{c}}{13}&{16}&{19}\\{14}&{17}&{20}\\{15}&{18}&{21}\end{array}\,} \right| = $

If $A = \left| {\,\begin{array}{*{20}{c}}{ - 1}&2&4\\3&1&0\\{ - 2}&4&2\end{array}\,} \right|$and $B = \left| {\,\begin{array}{*{20}{c}}{ - 2}&4&2\\6&2&0\\{ - 2}&4&8\end{array}\,} \right|$, then $B$ is given by

If ${a_1},{a_2},{a_3}.....{a_n}....$ are in $G.P.$ then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}\,} \right|$ is

$x + ky - z = 0,3x - ky - z = 0$ and $x - 3y + z = 0$ has non-zero solution for $k =$

If ${D_p} = \left| {\,\begin{array}{*{20}{c}}p&{15}&8\\{{p^2}}&{35}&9\\{{p^3}}&{25}&{10}\end{array}\,} \right|$, then ${D_1} + {D_2} + {D_3} + {D_4} + {D_5} = $