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?

If $\pi / 3$ is the angle between the straight lines $px + qy + r = 0$ and $x \sin \alpha + y \cos \alpha = r$ $(r \neq 0)$ which meet at a point $A$,and the straight line $x \cos \alpha - y \sin \alpha = 0$ also passes through the point $A$,then:

If the system of equations $2x + 3y = -1$,$3x + y = 2$,and $\lambda x + 2y = \mu$ is consistent,then:

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