The equation of the line passing through the point of intersection of the lines $x-3y+2=0$ and $2x+5y-7=0$ and perpendicular to the line $3x+2y+5=0$ is:

The lines $15x - 18y + 1 = 0$,$12x + 10y - 3 = 0$ and $6x + 66y - 11 = 0$ are

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

The line $(2 + k)x + (1 + k)y = 5 + 7k$ passes through a fixed point for all values of $k$. If $d$ is the distance of this fixed point from the origin, then $d^2 = \dots$

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:

$(a, b)$ is the point of concurrency of the lines $x-3y+3=0$,$kx+y+k=0$,and $2x+y-8=0$. If the perpendicular distance from the origin to the line $L \equiv ax-by+2k=0$ is $p$,then the perpendicular distance from the point $(2, 3)$ to $L=0$ is