Q:

Line situated in plane, finding two unknowns

I need to find $\alpha$ and $\beta$ so that the line d is situated inside the plane $\pi$.
We are given the equation of the plane
$$\pi:x-2y-3z-2=0$$
and the equation of line
$$d:\frac {x-\alpha}{2}=\frac{y+1}{\beta}=\frac{z-2\alpha}{3}$$
Using these I was able to find the direction vector
$$V_d(2,\beta,3) $$
and the point that intersects line d
$$A(2,-1,2\alpha)$$
and the system of equations
$$
\left\{ 
\begin{array}{c}
x=2(x_B-x_A)+\alpha \\ 
y=\beta(y_B-y_A)-1 \\ 
z=3(z_B-z_A)+2\alpha
\end{array}
\right. 
$$
I tried finding more formulas to help me,but this is all I found,what am I missing?

A:

$(2,-1,2\alpha)$ is a solution of $x-2y-3z-2=0$. So you can find $\alpha$.
$(2,-1,2\alpha)-(2,\beta,3)=(0, -1-\beta, 2\alpha-3)$ is a solution of $x-2y-3z-2=0$. So you can find $\beta$.

