The point on the line $3x + y + 4 = 0$ which is equidistant from $(-5, 6)$ and $(3, 2)$ is

If ${x_1}, {x_2}, {x_3}$ and ${y_1}, {y_2}, {y_3}$ are both in $G$.$P$. with the same common ratio,then the points $({x_1}, {y_1}), ({x_2}, {y_2})$ and $({x_3}, {y_3})$:

Consider the family of lines $x(a + b) + y = 1$,where $a, b$ and $c$ are the roots of the equation $x^3 - 3x^2 + x + \lambda = 0$ such that $c \in [1, 2]$. If the given family of lines makes a triangle of area $A$ with the coordinate axes,then the maximum value of $A$ (in sq. units) will be:

The point $(4,1)$ undergoes the following transformations successively:
$(i)$ Reflection in the line $x-y=0$
(ii) Shifting through a distance of $2$ units along the positive $X$-axis
(iii) Projection on the $X$-axis
The coordinates of the point in its final position are

Let two straight lines drawn from the origin $O$ intersect the line $3x + 4y = 12$ at the points $P$ and $Q$ such that $\triangle OPQ$ is an isosceles triangle and $\angle POQ = 90^{\circ}$. If $l = OP^2 + PQ^2 + QO^2$,then the greatest integer less than or equal to $l$ is:

Match the following:

List-$I$	List-$II$
$A$. The equation of line passing through $(4,3)$ whose $X$-intercept is twice its $Y$-intercept	$I$. $x+y-2\sqrt{2}=0$
$B$. The equation of the line passing through the centroid and circumcentre of $\triangle ABC$ with vertices $A(1,1), B(3,3), C(6,-6)$	$II$. $7x+23y-8=0$
$C$. The equation of the line whose $X$-intercept is $(-3/5)$ and is perpendicular to $x-y+2=0$	$III$. $x+2y+\sqrt{2}=0$
$D$. The equation of the line whose distance from the origin is $2$ and the normal drawn from the origin makes an angle $45^{\circ}$ with the positive direction of $X$-axis	$IV$. $x+2y-10=0$
	$V$. $5x+5y+3=0$