When the coordinate axes are rotated by an angle $\tan^{-1}\left(\frac{3}{4}\right)$ about the origin,then the equation $x^2+y^2=9$ is transformed to the equation

If $\theta_1, \theta_2, \theta_3$ are respectively the angles by which the coordinate axes are to be rotated to eliminate the $xy$ term from the following equations,then the descending order of these angles is:
$A_1 = 3x^2 + 5xy + 3y^2 + 2x + 3y + 4 = 0$
$A_2 = 5x^2 + 2\sqrt{3}xy + 3y^2 + 6 = 0$
$A_3 = 4x^2 + \sqrt{3}xy + 5y^2 - 4 = 0$

If $a \alpha^2+b \beta^2+c \alpha \beta+d=0$ is the transformed equation of $4 x^2+\sqrt{3} x y+5 y^2-4=0$ obtained by using $\alpha=\frac{\sqrt{3}}{2} x+\frac{y}{2}$ and $\beta=-\frac{x}{2}+\frac{\sqrt{3}}{2} y$,then $c(a+b+d)=$

Line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle keeping the origin fixed,the same line $L$ has intercepts $p$ and $q$. Then:

Let $P$ be the point to which the origin is shifted by the translation of axes so as to remove the first-degree terms from the equation $3x^2+y^2-6x+4y+4=0$. If the origin is shifted to $P$ by the translation of axes,then the transformed equation of $2x^2+3xy-5y^2+2x-23y-24=0$ is

The point $P(3,2)$ undergoes the following transformations successively:
$(i)$ Reflection about the line $y=x$
(ii) Translation to a distance of $3$ units in the positive direction of $X$-axis
(iii) Rotation through an angle $\frac{\pi}{4}$ about the origin in the counter-clockwise direction
Then,the final position of that point is