Which of the following lines is concurrent with the lines $3x + 4y + 6 = 0$ and $6x + 5y + 9 = 0$?

Let $a, b, c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4ax + 2ay + c = 0$ and $5bx + 2by + d = 0$ lies in the fourth quadrant and is equidistant from the two axes,then:

The lines $x-y-2=0$,$x+y-4=0$,and $x+3y=6$ meet at a common point:

The line passing through the intersection of the lines $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$ (where $(a, b) \neq (0, 0)$) and parallel to the $x$-axis is:

If three lines whose equations are $y=m_{1}x+c_{1}$,$y=m_{2}x+c_{2}$,and $y=m_{3}x+c_{3}$ are concurrent,then show that $m_{1}(c_{2}-c_{3})+m_{2}(c_{3}-c_{1})+m_{3}(c_{1}-c_{2})=0$.

$(x_1, y_1)$ is the point of concurrency of a family of lines. If the algebraic sum of the lengths of the perpendiculars drawn to these lines from $(2, 0)$,$(0, 2)$,and $(1, 1)$ is zero,then $(x_1, y_1) =$