The values of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6$,$3x+5y+5z=26$,and $x+2y+\lambda z=\mu$ has no solution are:

If the system of equations
$x+y+az=b$
$2x+5y+2z=6$
$x+2y+3z=3$
has infinitely many solutions,then $2a+3b$ is equal to $...........$.

If $A$ is a matrix such that $\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] A \left[\begin{array}{ll} 1 & 1 \end{array}\right] = \left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$, then $A$ is equal to

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:

If the system of linear equations $x+y+3z=0$,$x+3y+k^{2}z=0$,and $3x+y+3z=0$ has a non-zero solution $(x, y, z)$ for some $k \in R$,then $x + (y/z)$ is equal to

The sum of three numbers is $6$. Thrice the third number when added to the first number gives $7$. On adding three times the first number to the sum of the second and third numbers,we get $12$. The product of these numbers is: