If $\begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix} X = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}$,then $X =$

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 $S$ be the set of values of $\lambda$,for which the system of equations
$6 \lambda x - 3 y + 3 z = 4 \lambda^2$
$2 x + 6 \lambda y + 4 z = 1$
$3 x + 2 y + 3 \lambda z = \lambda$
has no solution. Then $12 \sum_{\lambda \in S} |\lambda|$ is equal to $...........$.

The positive value of $a$ for which the system of linear homogeneous equations $x+ay+z=0$,$ax+2y-z=0$,and $2x+3y+z=0$ has non-trivial solutions is

The system of linear equations $x + y + z = 6$,$2x + 5y + az = 36$,and $x + 2y + 3z = b$ has:

If the solution for the system of equations $x+2y-z=3$,$3x-y+2z=1$ and $2x-2y+3z=2$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2=$