If $a$ and $b$ are two arbitrary constants,then the straight line $(a - 2b)x + (a + 3b)y + 3a + 4b = 0$ will pass through

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

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

If the lines $ax + 2y + 1 = 0$,$bx + 3y + 1 = 0$,and $cx + 4y + 1 = 0$ are concurrent,then $a, b, c$ are in:

The lines $(a+2b)x + (a-3b)y = a-b$ for different values of $a$ and $b$ pass through a fixed point whose coordinates are:

If $\pi / 3$ is the angle between the straight lines $px + qy + r = 0$ and $x \sin \alpha + y \cos \alpha = r$ $(r \neq 0)$ which meet at a point $A$,and the straight line $x \cos \alpha - y \sin \alpha = 0$ also passes through the point $A$,then: