When the origin is shifted to the point (2, 3) and then the coordinate axes are rotated through an angle $\frac{\pi }{3}$ in the counter clockwise,
sense then the transformed equation of $3x^2 + 2xy + 3y^2 – 18x – 22y + 50 = 0$ is

$Given (x_1,y_1)=(2,3),\theta ={60}^0$

$$\begin{gathered} Given equation \\ 3x^2 + 2xy + 3y^2 – 18x – 22y + 50 = 0 \\ when the axes are translated to the point (2,3) then \\ transformed equation is 3x^2 + 2xy + 3y^2 +g{x}_1+f{y}_1+c=0 \\ \Rightarrow 3x^2 + 2xy + 3y^2-1=0\to \left(1\right) \\ Now axes are rotaed through an angle {60}^0 then \\ x=\frac{X-\sqrt{3}Y}{2} and y=\frac{\sqrt{3}X+Y}{2} \\ sub these values in \left(1\right) we get \\ 3{\left(\frac{X-\sqrt{3}Y}{2}\right)}^2+2\left(\frac{X-\sqrt{3}Y}{2}\right)\left(\frac{\sqrt{3}X+Y}{2}\right)+3{\left(\frac{\sqrt{3}X+Y}{2}\right)}^2-1=0 \\ \Rightarrow 6\sqrt{3}{X}^2-2XY+(6-2\sqrt{3})Y^2-2=0 \end{gathered}$$