If the system of equations $x + y+  z = 5$ ; $x + 2y + 3z = 9$ ; $x + 3y + \alpha z = \beta $ has infinitely many solutions, then $\beta  - \alpha $ equals

If ${x^a}{y^b} = {e^m},{x^c}{y^d} = {e^n},{\Delta _1} = \left| {\,\begin{array}{*{20}{c}}m&b\\n&d\end{array}\,} \right|\,\,{\Delta _2} = \left| {\,\begin{array}{*{20}{c}}a&m\\c&n\end{array}\,} \right|$ and ${\Delta _3} = \left| {\,\begin{array}{*{20}{c}}a&b\\c&d\end{array}\,} \right|$, then the values of $x$  and $y$  are respectively

If the system of linear equations $2 x+3 y-z=-2$  ; $x+y+z=4$  ; $x-y+|\lambda| z=4 \lambda-4$  (where $\lambda \in R$), has no solution, then

Evaluate the determinants

$\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$

Let $S_1$ and $S_2$ be respectively the sets of all $a \in R -\{0\}$ for which the system of linear equations

$a x+2 a y-3 a z=1$

$(2 a+1) x+(2 a+3) y+(a+1) z=2$

$(3 a+5) x+(a+5) y+(a+2) z=3$

has unique solution and infinitely many solutions. Then

In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$  has the value