$x+\sqrt{3} y=3$...(i)
$2 x^2+3 y^2=k$
Equation of tangent at $\mathrm{P}\left(x_1, y_1\right)$
$2 x x_1+3 y y_1=\mathrm{k}$...(ii)
If (i) and (ii) represents the same line
$$
\begin{aligned}
& \frac{1}{2 x_1}=\frac{\sqrt{3}}{3 y_1}=\frac{3}{k} \\
& x_1=\frac{k}{6}, y_1=\frac{k}{3 \sqrt{3}}
\end{aligned}
$$

Slope of tangent at $\mathrm{P}=\frac{-1}{\sqrt{3}}$
$\therefore$ Slope of normal $=\sqrt{3}$
$\therefore$ Equation of normal at $\mathrm{P}\left(x_1, y_1\right)$
$y-\frac{k}{3 \sqrt{3}}=\sqrt{3}\left(x-\frac{k}{6}\right) \Rightarrow \frac{3 \sqrt{3} y-k}{3}=\frac{6 x-k}{2}$...(iii)
Now, since $\left(x_1, y_1\right)=\left(\frac{k}{6}, \frac{k}{3 \sqrt{3}}\right)$ lies on ellipse
$$
\begin{aligned}
& \therefore 2\left(\frac{k}{6}\right)^2+3\left(\frac{k}{3 \sqrt{3}}\right)^2=k \\
& \frac{k^2}{18}+\frac{k^2}{9}=k \\
& \therefore k=6 \\
& \text { Put } k=6 \text { in }(3) \\
& 2(3 \sqrt{3} y-6)=3(6 x-6) \\
& \Rightarrow 3 x-\sqrt{3} y-1=0
\end{aligned}
$$