Coplanarity of two lines in 3D

Question

Suppose we have 2 lines $$l_1 : x = 5 , \frac{y}{3-\alpha}=\frac{z}{-2}$$ and $$ l_2: x= \alpha , \frac{y}{-1}= \frac{z}{2-\alpha}$$ so what will be value of $\alpha$ for lines to be coplaner ?

I believe that this is unsymmetrical form of lines equation and we have lines from intersecting planes in each of given equation , now I think for the lines to be coplanar , they should either be parallel or either should be intersecting hence one of the possible values of $\alpha$ should be $5$ but my answer does not matches and also for them being parallel $\frac{3-\alpha}{-1}=\frac{-2}{2-\alpha} $ but again my answer does not matches , help me a bit here

Answer 1

Both lines have constant $x$ values. If their $x$ values differ, then they can only be coplanar if the slopes in yz are equal. This is the equation you have set up, which gives $\alpha = 1 \mbox{ or } 4$

Another solution is given by forcing the two lines to both be in the plane $x = 5$, which comes by setting $\alpha$ to 5.

Answer 2

The parametric equations of $l_1$ are $$\left\{\begin{array}{l} x=5 \\ y=\lambda \\ z=-\frac{2}{3-\alpha}\lambda\end{array}\right.$$ and the parametric equations of $l_2$ are $$\left\{\begin{array}{l} x=\alpha \\ y=-\frac{1}{2-\alpha}\lambda \\ z=\lambda\end{array}\right.$$

Thus, we have $P_1(5,0,0)\in l_1,P_2(\alpha,0,0)\in l_2,$ from where $\vec{P_1P_2}=(\alpha-5,0,0),$ and $\vec{d}_1=\left(0,1,-\frac{2}{3-\alpha}\right),\vec{d}_2=\left(0,-\frac{1}{2-\alpha},1\right).$

Now, both lines are coplanar if they are parallel, that is,

$$\vec{d}_1\|\vec{d}_2 \iff \frac{1}{-\frac{2}{2-\alpha}}=-\frac{2}{3-\alpha} \iff (2-\alpha)(3-\alpha)=2 \iff\alpha\in \{1,4\},$$ or if they have a point in common, that is, $\alpha\ne 2,4$ and

$$\mathrm{ran}\left(\begin{array}{ccc} \alpha-5 & 0 & 0\\ 0 & 1 & -\frac{2}{3-\alpha}\\ 0 & -\frac{1}{2-\alpha} & 1 \end{array}\right)=2.$$ That is, $\alpha\ne 2,4$ and

$$0=\left|\begin{array}{ccc} \alpha-5 & 0 & 0\\ 0 & 1 & -\frac{2}{3-\alpha}\\ 0 & -\frac{1}{2-\alpha} & 1 \end{array}\right|=(\alpha-5)\left|\begin{array}{cc} 1& -\frac{2}{3-\alpha}\\ -\frac{1}{2-\alpha} & 1 \end{array}\right|\\ (\alpha-5)\left(1-\frac{2}{(3-\alpha)(2-\alpha)}\right)\iff \alpha=5,$$ since $\alpha\ne 2,4.$

So, the lines are parallel if $\alpha= 2,4$ and the lines have a point in common if $\alpha=5.$ These are the possible cases to be coplanar.