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

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 $x + y - z = 0,\,3x - \alpha y - 3z = 0,\,\,x - 3y + z = 0$ has non zero solution, then $\alpha = $

For all values of $A,B,C$ and $P,Q,R$, the value of $\left| {\,\begin{array}{*{20}{c}}{\cos (A - P)}&{\cos (A - Q)}&{\cos (A - R)}\\{\cos (B - P)}&{\cos (B - Q)}&{\cos (B - R)}\\{\cos (C - P)}&{\cos (C - Q)}&{\cos (C - R)}\end{array}\,} \right|$ is

The value of a for which the system of equations ${a^3}x + {(a + 1)^3}y + {(a + 2)^3}z = 0,$ $ax + (a + 1)y + (a + 2)z = 0,$ $x + y + z = 0,$ has a non zero solution is

Statement $-1 :$Determinant of a skew-symmetric matrix of order $3$ is zero

Statement $-2 :$ For any matrix $A,$ $\det \left( {{A^T}} \right) = {\rm{det}}\left( A \right)$ and $\det \left( { - A} \right) = - {\rm{det}}\left( A \right)$ Where $\det \left( A \right) = A$. Then :