The lines $x+2ay+a=0$,$x+3by+b=0$,and $x+4cy+c=0$ are concurrent. Then $a, b, c$ are in:

If $x+2y-3=0$,$3x+4y-7=0$,$2x+3y-4=0$,and $4x+5y-6=0$ are the equations of four lines,then

The equation of the line passing through the point of intersection of the lines $4x - 3y - 1 = 0$ and $5x - 2y - 3 = 0$ and parallel to the line $2y - 3x + 2 = 0$ is:

If $P \equiv \left( \frac{1}{x_p}, p \right), Q = \left( \frac{1}{x_q}, q \right), R = \left( \frac{1}{x_r}, r \right)$ where $x_k \neq 0$ denotes the $k^{th}$ term of an $H.P.$ for $k \in N$,then:

The equation of the line passing through the point of intersection of the lines $x + 2y + 6 = 0$ and $2x - y = 2$ and making an intercept $5$ on the $y$-axis is

The system of equations $2x + y - 5 = 0$,$x - 2y + 1 = 0$,and $2x - 14y - a = 0$ is consistent. Then,$a$ is equal to