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

If $A = \left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]$ and $\det ({A^n} - I) = 1 - {\lambda ^n}\,,\,n \in N$ then $\lambda $ is

Statement $1$ : If the system of equations $x + ky + 3z = 0, 3x+ ky - 2z = 0, 2x + 3y - 4z = 0$ has a nontrivial solution, then the value of $k$ is $\frac{31}{2}$

Statement $2$ : A system of three homogeneous equations in three variables has a non trivial solution if the determinant of the coefficient matrix is zero.

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

The values of $\mathrm{m}, \mathrm{n}$, for which the system of equations

$ x+y+z=4 $

$ 2 x+5 y+5 z=17 $

$ x+2 y+m z=n$

has infinitely many solutions, satisfy the equation :