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:

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

Let $a, b$ and $c$ be distinct and none of them is equal to $1$. If the lines $x+ay+a=0$,$bx+y+b=0$ and $cx+cy+1=0$ are concurrent,then the value of $\frac{a}{a-1}+\frac{b}{b-1}+\frac{c}{c-1}$ is

If the lines $4x + 3y - 1 = 0$,$x - y + 5 = 0$,and $kx + 5y - 3 = 0$ are concurrent,then $k=$

Given two points $Q(3,4)$ and $R(1,2)$. What is the point $P(x, y)$ on the line $2x-y-1=0$ for which $PQ+PR=QR$ holds?