If the lines $x + 2ay + a = 0$,$x + 3by + b = 0$ and $x + 4cy + c = 0$ are concurrent,then $a$,$b$ and $c$ are in

Consider the set of all lines $px + qy + r = 0$ such that $3p + 2q + 4r = 0$. Which one of the following statements is true?

What is the nature of the points $(-a, -b)$,$(0, 0)$,$(a, b)$,and $(a^2, ab)$?

Let $\alpha, \beta, \gamma$ be three non-zero real constants and $a, b, c$ be three arbitrary real numbers which satisfy $\alpha a + \beta b + \gamma c = 0$. Then the point of concurrence of the family of lines $ax + by + c = 0$ is

If the point of intersection of the lines $2ax + 4ay + c = 0$ and $7bx + 3by - d = 0$ lies in the $4^{th}$ quadrant and is equidistant from the two axes, where $a, b, c,$ and $d$ are non-zero numbers, then $ad : bc$ is equal to

If $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$, then the lines $a_i x + b_i y + c_i = 0$ $(i = 1, 2, 3)$ represent: