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?

Statement $(A)$: If $3a - 2b + 5c = 0$,then the line $ax + by + c = 0$ is always concurrent at a point.
Reason $(R)$: If $L_1 = 0$ and $L_2 = 0$ are two lines,then the family of lines $L_1 + \lambda L_2 = 0$ is concurrent at the intersection of $L_1$ and $L_2$.

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

For the straight lines given by the equation $(2 + k)x + (1 + k)y = 5 + 7k$,for different values of $k$,which of the following statements is true?

Find the value of $p$ so that the three lines $3x + y - 2 = 0$,$px + 2y - 3 = 0$,and $2x - y - 3 = 0$ intersect at a single point.

$(a, b)$ 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