The system of equations $x+3by+bz=0$, $x+2ay+az=0$ and $x+4cy+cz=0$ has

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 $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 0 & 5 \\ 0 & 2 & 1 \end{bmatrix}$ and $b = \begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}$. Which of the following is true?

If the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$,and $x + y + 3z = 4$ has infinitely many solutions,then $p=$

$A, C$ are $3 \times 3$ matrices. $B, D$ are $3 \times 1$ matrices. If $AX=B$ has a unique solution and $CX=D$ has an infinite number of solutions, 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 $\left|\begin{array}{ccc}1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4\end{array}\right| = 0$.