The point $P$ is equidistant from $A(1, 3)$,$B(-3, 5)$,and $C(5, -1)$. Then $PA$ is equal to:

If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos \theta + y \sin \theta = 7, \theta \in \left(0, \frac{\pi}{2}\right)$ between the coordinate axes,then $\alpha$ is equal to

$A(5,3), B(3,-2), C(2,-1)$ are three points. If $P(x,y)$ is a variable point such that the area of the quadrilateral $PABC$ is $10$ sq. units,then the locus of $P$ is

If $a, b, c$ are in Arithmetic Progression $(AP)$,then the line $ax + by + c = 0$ always passes through a fixed point. The coordinates of this point are:

If $A(2, -3)$ and $B(-2, 1)$ are two vertices of a $\triangle ABC$ and if the centroid of $\triangle ABC$ lies on the line $2x + 3y = 1$,then the locus of vertex $C$ of $\triangle ABC$ is equal to

$A$ variable line passes through a fixed point $P$. If the algebraic sum of the perpendiculars drawn from $(2, 0)$,$(0, 2)$,and $(1, 1)$ to the line is zero,then the coordinates of the point $P$ are: