If the system of linear equations given by $x+y+z=3$,$2x+2y-z=3$,and $x+y-z=1$ is consistent and if $(x_0, y_0, z_0)$ is a solution,then $2x_0+2y_0+z_0=$

The solution of equations $x + y = 10$,$2x + y = 18$,and $4x - 3y = 26$ is:

If $\begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 12 \\ 15 \\ 13 \end{bmatrix}$,then the value of $x^2 + y^2 + z^2 =$

The system of equations $x+y+z=6$,$x+2y+5z=9$,$x+5y+\lambda z=\mu$ has no solution if

Consider the system of equations:
$x - 2y + 3z = -1$; $-x + y - 2z = k$; $x - 3y + 4z = 1$
$STATEMENT-1$: The system of equations has no solution for $k \neq 3$.
$STATEMENT-2$: The determinant $\left|\begin{array}{ccc}1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4\end{array}\right| = 0$.

Let $A = \begin{bmatrix} 12 & 24 & 5 \\ x & 6 & 2 \\ -1 & -2 & 3 \end{bmatrix}$. The value of $x$ for which the matrix $A$ is not invertible is