What is the ratio of the area of $\triangle PQS$ to the area of $\triangle PQR$?

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:

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

The perimeter of a triangle is $16 \text{ cm}$, one of the sides is of length $6 \text{ cm}$. If the area of the triangle is $12 \text{ cm}^2$, then the triangle is:

The point $P(3,6)$ is first reflected on the line $y=x$ and then the image point $Q$ is again reflected on the line $y=-x$ to get the image point $Q^{\prime}$. Then, the circumcentre of the $\Delta P Q Q^{\prime}$ is

If the coordinates of points $A, B, C$ are $(-1, 5), (0, 0)$ and $(2, 2)$ respectively,and $D$ is the midpoint of $BC$,then the equation of the perpendicular drawn from $B$ to the line $AD$ is:

$A$ ray of light coming from the point $P(1, 2)$ gets reflected from the point $Q$ on the $x$-axis and then passes through the point $R(4, 3)$. If the point $S(h, k)$ is such that $PQRS$ is a parallelogram,then $hk^2$ is equal to :