If the system of equations $x+ky+3z=-2$, $4x+3y+kz=14$, and $2x+y+2z=3$ can be solved by the matrix inversion method, then:

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 $\left|\begin{array}{ccc}1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4\end{array}\right| = 0$.

If the system of equations $x+y+2z=3$, $x+2y+3z=4$ and $x+y+cz=5$ is inconsistent, then:

For the system $x-y+z=4, 2x+y-3z=0, x+y+z=2$,the values of $x, y, z$ respectively are given by

Let $A = \begin{bmatrix} 1 & -4 & 7 \\ 0 & 3 & -5 \\ -2 & 5 & -9 \end{bmatrix}$ and $B = \begin{bmatrix} a \\ -b \\ -c \end{bmatrix}$. If $A$ and $[A: B]$ have the same rank, then:

The system of equations $kx + y + z = 1$,$x + ky + z = k$,and $x + y + kz = k^2$ has no solution if $k$ is equal to