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

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

The point of intersection of the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{b} + \frac{y}{a} = 1$ lies on the line

If $(x_1, y_1)$ are the roots of $x^2 + 8x - 20 = 0$,$(x_2, y_2)$ are the roots of $4x^2 + 32x - 57 = 0$ and $(x_3, y_3)$ are the roots of $9x^2 + 72x - 112 = 0$,then the points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$:

If the points $(k, 2 - 2k)$,$(1 - k, 2k)$ and $(-k - 4, 6 - 2k)$ are collinear,then the possible values of $k$ are