Solve the system of linear equations using the matrix method: $2x - y = -2$ and $3x + 4y = 3$.

If the system of equations
$ 11 x+y+\lambda z=-5 $
$ 2 x+3 y+5 z=3 $
$ 8 x-19 y-39 z=\mu $
has infinitely many solutions,then $ \lambda^4-\mu $ is equal to :

The number of solutions of the equations $x + 4y - z = 0,$ $3x - 4y - z = 0,$ and $x - 3y + z = 0$ is

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 $a = -3$.

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 $x^{2}+y^{2}+z^{2}=$

Solve the system of linear equations using the matrix method: $x-y+2z=7$,$3x+4y-5z=-5$,$2x-y+3z=12$.