If the system of equations $\alpha x+y+z=5, x+2 y+$ $3 z=4, x+3 y+5 z=\beta$ has infinitely many solutions, then the ordered pair $(\alpha, \beta)$ is equal to:

If the system of linear equations $2 \mathrm{x}+2 \mathrm{ay}+\mathrm{az}=0$ ; $2 x+3 b y+b z=0$ ; $2 \mathrm{x}+4 \mathrm{cy}+\mathrm{cz}=0$ ; where $a, b, c \in R$ are non-zero and distinct; has a non-zero solution, then 

If $\left| {{\kern 1pt} \begin{array}{*{20}{c}}1&2&3\\2&x&3\\3&4&5\end{array}\,} \right| = 0,$ then  $x =$

If ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right|$ and ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}x&b\\a&x\end{array}\,} \right|$ are the given determinants, then

Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$, and a system of linear equations

$x+y+z=5$    ;    $x+2 y+3 z=\mu$   ;     $x+3 y+\lambda z=1$

is constructed. If $\mathrm{p}$ is the probability that the system has a unique solution and $\mathrm{q}$ is the probability that the system has no solution, then :

If $p{\lambda ^4} + q{\lambda ^3} + r{\lambda ^2} + s\lambda + t = $ $\left| {\,\begin{array}{*{20}{c}}{{\lambda ^2} + 3\lambda }&{\lambda - 1}&{\lambda + 3}\\{\lambda + 1}&{2 - \lambda }&{\lambda - 4}\\{\lambda - 3}&{\lambda + 4}&{3\lambda }\end{array}\,} \right|,$ the value of $t$ is