The system of linear equations $x + y + z = 2$,$2x + y - z = 3$,and $3x + 2y + kz = 4$ has a unique solution if

Consider the system of equations in $x, y$ and $z$:
$12x + by + cz = 0$
$ax + 24y + cz = 0$
$ax + by + 36z = 0$
(where $a, b, c$ are real numbers,$a \ne 12, b \ne 24, c \ne 36$).
If the system of equations has a non-trivial solution $(z \ne 0)$,then the value of $\frac{1}{a - 12} + \frac{2}{b - 24} + \frac{3}{c - 36}$ is:

If the system of equations $2x + \lambda y + 3z = 5$,$3x + 2y - z = 7$,and $4x + 5y + \mu z = 9$ has infinitely many solutions,then $(\lambda^2 + \mu^2)$ is equal to:

Let the system of linear equations $x + 2y + z = 2$,$\alpha x + 3y - z = \alpha$,and $-\alpha x + y + 2z = -\alpha$ be inconsistent. Then $\alpha$ is equal to:

The product of all real values of $b$ such that there is no solution to the system of equations $2x + 5y + z = 19$,$-4x + by + 6z = -42$,and $-3y - bz = 81$ is:

Let $n$ be the number obtained on rolling a fair die. If the probability that the system of equations
$x-ny+z=6$
$x+(n-2)y+(n+1)z=8$
$(n-1)y+z=1$
has a unique solution is $\frac{k}{6}$,then the sum of $k$ and all possible values of $n$ is: