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:

The value of $\lambda$ for which the system of equations $2x - y - z = 12,$ $x - 2y + z = -4,$ and $x + y + \lambda z = 4$ has no solution is

In a legislative assembly election,a political group hired a public relations firm to promote its candidate in three ways: telephone,house calls,and letters. The cost per contact (in paise) is given in matrix $A$ as $A = \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix} \begin{matrix} \text{Telephone} \\ \text{Housecall} \\ \text{Letter} \end{matrix}$. The number of contacts of each type made in two cities $X$ and $Y$ is given by $B = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{matrix} \text{Telephone} & \text{Housecall} & \text{Letter} \\ \to X \\ \to Y \end{matrix}$. Find the total amount spent by the group in the two cities $X$ and $Y$.

If the system of equations $kx + (k+1)y + (k-1)z = 0$,$(k-1)x + (k+2)y + kz = 0$,and $(k+1)x + ky + (k+2)z = 0$ has a non-trivial solution,then the sum of all possible values of $k$ is:

If the solution for the system of equations $x+2y-z=3$,$3x-y+2z=1$ and $2x-2y+3z=2$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2=$

The system of equations $x+3by+bz=0$, $x+2ay+az=0$ and $x+4cy+cz=0$ has