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 the lines $x+3y-5=0$,$5x+2y-12=0$,and $3x-ky-1=0$ do not form a triangle,then a value of $k$ is

The point of intersection of the lines joining points $\hat{i}+2 \hat{j}, 2 \hat{i}-\hat{j}$ and $-\hat{i}, 2 \hat{i}$ is

The equation of the line passing through the point of intersection of the lines $2x + 3y + 6 = 0$ and $3x - y - 13 = 0$ and parallel to the line $3x - 4y + 5 = 0$ is

The line on which the lines $ax + by = 1$ and $bx + ay = 1$ (with $a \neq 0 \neq b$) intersect for any real values of $a$ and $b$ is

If the three distinct lines $x + 2ay + a = 0$,$x + 3by + b = 0$,and $x + 4ay + a = 0$ are concurrent,then the point $(a, b)$ lies on a