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

Consider the system of equations: $ax + by + cz = 2$, $bx + cy + az = 2$, $cx + ay + bz = 2$, where $a, b, c$ are real numbers such that $a + b + c = 0$. Then, the system

Solve the system of linear equations using the matrix method:
$5x + 2y = 3$
$3x + 2y = 5$

Let for any three distinct consecutive terms $a, b, c$ of an $A.P.$,the lines $ax + by + c = 0$ be concurrent at the point $P$ and $Q(\alpha, \beta)$ be a point such that the system of equations $x + y + z = 6$,$2x + 5y + \alpha z = \beta$ and $x + 2y + 3z = 4$ has infinitely many solutions. Then $(PQ)^2$ is equal to . . . . . . .

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:

Let $[\lambda]$ be the greatest integer less than or equal to $\lambda$. The set of all values of $\lambda$ for which the system of linear equations $x+y+z=4$,$3x+2y+5z=3$,$9x+4y+(28+[\lambda])z=[\lambda]$ has a solution is: