Which of the following lines is concurrent with the lines $3x + 4y + 6 = 0$ and $6x + 5y + 9 = 0$?

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:

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

Consider the lines given by $L_1: x+3y-5=0$,$L_2: 3x-ky-1=0$,and $L_3: 5x+2y-12=0$. Match the statements in Column $I$ with the statements in Column $II$.

Column $I$	Column $II$
$(A)$ $L_1, L_2, L_3$ are concurrent,if	$(p)$ $k=-9$
$(B)$ One of $L_1, L_2, L_3$ is parallel to at least one of the other two,if	$(q)$ $k=-\frac{6}{5}$
$(C)$ $L_1, L_2, L_3$ form a triangle,if	$(r)$ $k=\frac{5}{6}$
$(D)$ $L_1, L_2, L_3$ do not form a triangle,if	$(s)$ $k=5$

The lines $x-2y+1=0$,$2x-3y-1=0$,and $3x-y+k=0$ are concurrent. The angle between the lines $3x-y+k=0$ and $mx-3y+6=0$ is $45^{\circ}$. If $m$ is an integer,then $m-k=$

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