If the system of equations $x+4y-z=\lambda$,$7x+9y+\mu z=-3$,and $5x+y+2z=-1$ has infinitely many solutions,then $(2\mu+3\lambda)$ is equal to:

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

Let $[\lambda]$ be the greatest integer less than or equal to $\lambda$. The set of all values of $\lambda$ for which the system of linear equations $x+y+z=4$,$3x+2y+5z=3$,$9x+4y+(28+[\lambda])z=[\lambda]$ has a solution is:

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:

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:

In solving a system of linear equations $AX=B$ by Cramer's rule,in the usual notation,if $\Delta_1=\left|\begin{array}{ccc}-11 & 1 & -7 \\ -4 & 1 & -2 \\ 5 & 1 & 1\end{array}\right|$ and $\Delta_3=\left|\begin{array}{ccc}4 & 1 & -11 \\ 1 & 1 & -4 \\ 4 & 1 & 5\end{array}\right|$,then $X=$