An ordered pair $(\alpha , \beta )$ for which the system of linear equations

$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x  + \beta y + 2z = 2$ has a unique solution, is

If the system of linear equations $x+ ay+z\,= 3$ ; $x + 2y+ 2z\, = 6$ ; $x+5y+ 3z\, = b$ has no solution, then

The roots of the determinant equation (in $x$) $\left| {\,\begin{array}{*{20}{c}}a&a&x\\m&m&m\\b&x&b\end{array}\,} \right| = 0$

If $n \ne 3k$ and 1, $\omega ,{\omega ^2}$ are the cube roots of unity, then $\Delta = \left| {\,\begin{array}{*{20}{c}}1&{{\omega ^n}}&{{\omega ^{2n}}}\\{{\omega ^{2n}}}&1&{{\omega ^n}}\\{{\omega ^n}}&{{\omega ^{2n}}}&1\end{array}\,} \right|$ has the value

If $a \ne p,b \ne q,c \ne r$ and $\left| {\,\begin{array}{*{20}{c}}p&b&c\\{p + a}&{q + b}&{2c}\\a&b&r\end{array}\,} \right|$ =$ 0$, then $\frac{p}{{p – a}} + \frac{q}{{q – b}} + \frac{r}{{r – c}} = $