For all values of $a$ and $b$,the line $(a+2b)x + (a-b)y + (a+5b) = 0$ passes through a fixed point. Find that point.

If three non-zero real numbers $a, b, c$ are in harmonic progression,then the straight lines $\frac{x}{a} + \frac{y}{b} - \frac{2}{c} = 0$ are concurrent at the point

The lines $x-2y+1=0$,$2x-3y-1=0$,and $3x-y+k=0$ are concurrent. The angle between the lines $3x-y+k=0$ and $mx-3y+6=0$ is $45^{\circ}$. If $m$ is an integer,then $m-k=$

The line $(2 + k)x + (1 + k)y = 5 + 7k$ passes through a fixed point for all values of $k$. If $d$ is the distance of this fixed point from the origin, then $d^2 = \dots$

The number of values of $a$ for which the system of equations $a^2 x + (2 - a) y = 4 + a^2$ and $a x + (2 a - 1) y = a^5 - 2$ possesses no solution is:

Number of values of $m$ for which the lines $x + y - 1 = 0$,$(m - 1)x + (m^2 - 7)y - 5 = 0$ and $(m - 2)x + (2m - 5)y = 0$ are concurrent,are