Let the system of linear equations $x+y+kz=2$; $2x+3y-z=1$; $3x+4y+2z=k$ have infinitely many solutions. Then the system $(k+1)x+(2k-1)y=7$; $(2k+1)x+(k+5)y=10$ has:

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 linear equations $x + ky + 3z = 0$,$3x + ky - 2z = 0$,and $2x + 4y - 3z = 0$ has a non-zero solution $(x, y, z)$,then $\frac{xz}{y^2} = \dots$

For how many values of $k$ does the system of linear equations $(k + 1)x + 8y = 4k$ and $kx + (k + 3)y = 3k - 1$ have no solutions?

If the system of linear equations $2x - 3y = \gamma + 5$ and $\alpha x + 5y = \beta + 1$,where $\alpha, \beta, \gamma \in R$,has infinitely many solutions,then the value of $|9\alpha + 3\beta + 5\gamma|$ is equal to

Let $a, \lambda, \mu \in \mathbb{R}$. Consider the system of linear equations:
$a x + 2 y = \lambda$
$3 x - 2 y = \mu$
Which of the following statement$(s)$ is(are) correct?
$(A)$ If $a = -3$,then the system has infinitely many solutions for all values of $\lambda$ and $\mu$.
$(B)$ If $a \neq -3$,then the system has a unique solution for all values of $\lambda$ and $\mu$.
$(C)$ If $\lambda + \mu = 0$,then the system has infinitely many solutions for $a = -3$.
$(D)$ If $\lambda + \mu \neq 0$,then the system has no solution for $a = -3$.