If $AX=B$,where $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 2 & 1\end{array}\right]$,$X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$,and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$,then find the value of $2x+y-z$.

Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$,and a system of linear equations
$x+y+z=5$
$x+2y+3z=\mu$
$x+3y+\lambda z=1$
is constructed. If $p$ is the probability that the system has a unique solution and $q$ is the probability that the system has no solution,then:

Find the matrix $X$ such that $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.

If $3X + 2Y = I$ and $2X - Y = O$,where $I$ and $O$ are unit and null matrices of order $3$ respectively,then

If $AX=B$,where $A=\begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4 \end{bmatrix}$,$X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $B=\begin{bmatrix} 12 \\ 15 \\ 13 \end{bmatrix}$,then $x^{2}+y^{2}+z^{2}=$

The system of equations $x+y+z=5$, $x+2y+3z=9$ and $x+3y+\lambda z=\mu$ has a unique solution if