If the incentre of the triangle formed by the lines $x-2=0$,$x+y-1=0$,and $x-y+3=0$ is $(\alpha, \beta)$,then $\beta=$

In the triangle with vertices at $A(6,3), B(-6,3)$ and $C(-6,-3)$,the median through $A$ meets $BC$ at $P$,the line $AC$ meets the $x$-axis at $Q$,while $R$ and $S$ respectively denote the orthocentre and centroid of the triangle. Then the correct matching of the coordinates of points in List-$I$ to List-$II$ is:

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

If the vertices of a triangle $ABC$ are $A(1,7)$,$B(-5,-1)$,and $C(7,4)$,then the equation of the internal angle bisector of $\angle ABC$ is

Let the mirror image of point $A(\alpha, \beta)$ in the line mirror $x + 2y = 3$ be point $B$,and the image of $B$ in the line $3x - 2y = 5$ be $C$. If the origin is the orthocentre of triangle $ABC$ and $P(a, b)$ is a point inside the triangle such that triangles $PAB$,$PBC$,and $PCA$ have the same area,then $3(a + b)$ is:

Let $PQR$ be a triangle in which $PQ=3$. From the vertex $R$,draw the altitude $RS$ to meet $PQ$ at $S$. Assume that $RS=\sqrt{3}$ and $PS=QR$. Then,$PR$ equals

If the points $(0, 0)$,$(2, 2\sqrt{3})$ and $(a, b)$ are the vertices of an equilateral triangle,then $(a, b) = $