The equation of the line passing through the intersection of the lines $x - y = 4$ and $3x + y = 7$ and parallel to the line $x + 2y = 1$ is:

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 joining points $\hat{i}+2 \hat{j}, 2 \hat{i}-\hat{j}$ and $-\hat{i}, 2 \hat{i}$ is

For $a, b, c \in R$,if $6 a^2-3 b^2-c^2+7 a b-a c+4 b c=0$ and $|a|+|b| \neq 0$,then all the lines given by $a x+b y+c=0$ are

If three lines whose equations are $y=m_{1}x+c_{1}$,$y=m_{2}x+c_{2}$,and $y=m_{3}x+c_{3}$ are concurrent,then show that $m_{1}(c_{2}-c_{3})+m_{2}(c_{3}-c_{1})+m_{3}(c_{1}-c_{2})=0$.

Given the four lines with equations $x + 2y = 3,$ $3x + 4y = 7,$ $2x + 3y = 4,$ and $4x + 5y = 6,$ these lines are: