For how many different values of $a$ does the following system of linear equations have at least two distinct solutions?
$ax + y = 0$
$x + (a + 10)y = 0$

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 $x+y+z=$

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:

The equations $x + 2y + 3z = 1,$ $2x + y + 3z = 2,$ and $5x + 5y + 9z = 4$ have:

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 . . . . . . .

The number of real values of $\lambda$ for which the system of linear equations $2x + 4y - \lambda z = 0$,$4x + \lambda y + 2z = 0$,and $\lambda x + 2y + 2z = 0$ has infinitely many solutions is: