The value of $\lambda$ such that the system of equations $2x-y-2z=2$, $x-2y+z=-4$, and $x+y+\lambda z=4$ has no solution, is:

Let $A$ be a $2 \times 2$ matrix with $\det(A)=-1$ and $\det((A+I)(\operatorname{Adj}(A)+I))=4$. Then the sum of the diagonal elements of $A$ can be.

Let $S$ be the set of values of $\lambda$,for which the system of equations
$6 \lambda x - 3 y + 3 z = 4 \lambda^2$
$2 x + 6 \lambda y + 4 z = 1$
$3 x + 2 y + 3 \lambda z = \lambda$
has no solution. Then $12 \sum_{\lambda \in S} |\lambda|$ is equal to $...........$.

If $AX=D$ represents the system of linear equations $3x-4y+7z+6=0$, $5x+2y-4z+9=0$ and $8x-6y-z+5=0$, then

For what values of $a$ will the system of equations $x+y+z=1$, $2x+3y+2z=2$, and $ax+ay+2az=4$ have a unique solution?

The values of $p$ and $q$ so that the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$ and $x + y + 3z = 4$ may have no solution are