Family of Straight Line

Since $a, 2b, 3c$ are in an arithmetic progression, let $a = 2b-d$ and $3c = 2b + d$ , where $d$ is the common difference. Then, we have:

$\begin{aligned} 2ax + 2by + c & = 0 \end{aligned}$

$\begin{aligned} y & = - \frac {2a}{2b}x - \frac c{2b} \\ & = - \frac {2b-d}b x - \frac {2b+d}{6b} \\ & = \left(\frac db - 2\right) x - \frac 13 - \frac d{6b} \end{aligned}$

Therefore, $2ax + 2by + c = 0$ is a family of straight lines with $\dfrac db$ as parameter. We note that when $x=\dfrac 16$ , $y = \left(\dfrac db - 2\right) \dfrac 16 - \dfrac 13 - \dfrac d{6b} = \dfrac d{6b} - \dfrac 13 - \dfrac 13 - \dfrac d{6b} = - \dfrac 23$ , which is independent of $d$ and $b$ . Therefore, the lines are concurrent at $\boxed{\left(\dfrac 16, -\dfrac 23\right)}$ .

Let $a, 2b = a+d, 3c = a + 2d$ be in arithmetic progression. If $2ax+2by+c=0$ represents a family of lines concurrent about a fixed point, then we have:

$2ax + (a+d)y = -\frac{a+2d}{3};$

or $6ax + 3(a+d)y = -a-2d;$

or $(6x + 3y)a + (3y)d = -a - 2d;$

If we now match the coefficients of $a$ and $d$ , then we arrive at the 2x2 linear system:

$6x+3y=-1$

$3y = -2$

which has the unique solution $\boxed{(x,y) = (\frac{1}{6},-\frac{2}{3})}.$