Given line $2 x-3 y+4=0$ cuts the ellipse $x=3 \cos$ $\theta, y=5 \sin \theta$.
Here, $a=3, b=5$.
Required equation of ellipse is
$$
\begin{aligned}
& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\
& \frac{x^2}{9}+\frac{y^2}{25}=1 . .
\end{aligned}
$$
Satisfy the line $3 y=2 x+4$
$$
\mathrm{y}=\frac{2 \mathrm{x}+4}{3}
$$
From (i)
$$
\begin{aligned}
& 25 \mathrm{x}^2+9 \mathrm{y}^2=225 \\
& 25 \mathrm{x}^2+9\left(\frac{2 \mathrm{x}+4}{3}\right)^2=225 \\
& 25 \mathrm{x}^2+\frac{9}{9}\left(4 \mathrm{x}^2+16+16 \mathrm{x}\right)=225 \\
& 29 \mathrm{x}^2+16 \mathrm{x}-209=0
\end{aligned}
$$
Let $\mathrm{A}$ is $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$
$$
\begin{aligned}
& \mathrm{x}_1+\mathrm{x}_2=\frac{-16}{29} \\
& \frac{\mathrm{x}_1+\mathrm{x}_2}{2}=\frac{-8}{29}
\end{aligned}
$$
Here, $(\alpha, \beta)$ is the mid point of the line $\mathrm{AB}$
$$
\alpha=\frac{-8}{29}
$$
Now, $2 x=3 y-4 \Rightarrow x=\frac{3 y-4}{2}$
From (i)
$$
\begin{aligned}
& \frac{25}{4}\left(16+9 y^2-24 y\right)+9 y^2=225 \\
& 225 y^2+36 y^2-600 y+400=900
\end{aligned}
$$
$$
261 y^2-600 y-500=0
$$
So, $\mathrm{y}_1+\mathrm{y}_2=\frac{600}{261}$
$$
\begin{aligned}
& \frac{\mathrm{y}_1+\mathrm{y}_2}{2}=\frac{300}{261} \\
& \beta=\frac{300}{261}
\end{aligned}
$$
Now, $3 \beta-2 \alpha=3 \times \frac{300}{261}+2 \times \frac{8}{29}$
$$
=\frac{300}{87}+\frac{16}{29}=\frac{300+48}{87}=\frac{348}{87}=4
$$
So, option (b) is correct.