Given hyperbola is 3x² - 4y² =12
$\begin{array}{l}\Rightarrow \frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{3}=1\\ \Rightarrow {\mathrm{a}}^2=4,{\mathrm{b}}^2=3\end{array}$
$\therefore$Equation of normal at P($\mathrm{\theta }$) is 
$\begin{array}{l}\frac{\mathrm{ax}}{\sec \mathrm{\theta }}+\frac{\mathrm{by}}{\tan \mathrm{\theta }}={\mathrm{a}}^2+{\mathrm{b}}^2\\ \Rightarrow \frac{2\mathrm{x}}{\sec \frac{\mathrm{\pi }}{3}}+\frac{\sqrt{3}\mathrm{y}}{\tan \frac{\mathrm{\pi }}{3}}=4+3\\ \Rightarrow \frac{1}{2}\left(2\mathrm{x}\right)+\frac{1}{\sqrt{3}}\left(\sqrt{3}\mathrm{y}\right)=7\\ \Rightarrow \mathrm{x}+\mathrm{y}=7\end{array}$