Line situated in plane, finding two unknowns

Question

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?

Answer 1

$(2,-1,2\alpha)$ 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 $\beta$.

Answer 2

Any point on the line is of the form $(\alpha +2t,-1+\beta t,2\alpha +3t)$ where $t$ (the common ratio) is a real number. Just plug these values in the equation for the plane. In order that the resulting equation holds for all $t$ the coefficient of $t$ and the constant term must both be $0$. This gives you the values of $\alpha $ and $\beta $.

Answer 3

The direction vector of the line must be orthogonal to the plane’s normal. Both of these can be read directly from the given equations, yielding the equation $$(2,\beta,3)\cdot(1,-2,-3)=0,$$ which you can easily solve for $\beta$.

You can also read a point on the line directly from its equations: $(\alpha,-1,2\alpha)$. Plugging this point into the equation of the plane gives you a simple equation in $\alpha$ to solve.