All the real values of $p, q$ so that the system of equations $\begin{cases} 2x + py + 6z = 8 \\ x + 2y + qz = 5 \\ x + y + 3z = 4 \end{cases}$ may have no solution are

The number of solutions of the equations $x + y - z = 0$,$3x - y - z = 0$,and $x - 3y + z = 0$ is

Let $AX=D$ be a system of three linear non-homogeneous equations. If $|A|=0$ and $\operatorname{rank}(A)=\operatorname{rank}([AD])=\alpha$, then

If the system of equations
$2x + 7y + \lambda z = 3$
$3x + 2y + 5z = 4$
$x + \mu y + 32z = -1$
has infinitely many solutions,then $(\lambda - \mu)$ is equal to

Statement $1$: If the system of equations $x + ky + 3z = 0, 3x + ky - 2z = 0, 2x + 3y - 4z = 0$ has a nontrivial solution,then the value of $k$ is $\frac{31}{2}$.
Statement $2$: $A$ system of three homogeneous equations in three variables has a nontrivial solution if the determinant of the coefficient matrix is zero.

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 . . . . . . .