Find the value of $a$ for which the two lines are on the same plane Find the value(s) of $a$ for which the two lines $$L_1:=\left\{\left(-1,0,-1\right)+t\left(a,2,0\right):t \in \mathbb R\right\}$$
$$L_2:=\left\{\left(1,2a,1\right)+t\left(2,3,2\right):t \in \mathbb R\right\}$$
place on the same plane in $\mathbb R^3$.

The equation of a plane is :
$$A\left(x-x_{0}\right)+B\left(y-y_{0}\right)+C\left(z-z_{0}\right)=0$$
Where $n=\left(A,B,C\right)$ is normal to the plane and $P=(x_0,y_0,z_0)$ is on the plane.
Since the two lines are going to be on the plane,then we conclude that the normal to the lines are also normal to the plane.
The parametric equation of the lines is:$$L_1:x=at-1,y=2t,z=-1$$$$L_2:x=2t+1,y=3t+2a,z=2t+1$$
The points $P(a-1,2,-1)$ and $Q(-1,0,-1)$ are on $L_1$ and the point $U(1,2a,1)$ is on $L_2$,we can construct the normal to the plane by computing $n=\vec{ UP} \times \vec{UQ}=(a-2,2-2a,-2) \times (-2,-2a,-2)=(-4,2a,-2a^2+4)$
Since the two points $P,Q$ are on the plane we see that:
$$-4\left(x+1\right)+2a\left(y\right)+\left(-2a^{2}+4\right)\left(z+1\right)=0$$$$-4\left(x-1\right)+2a\left(y-2a\right)+\left(-2a^{2}+4\right)\left(z-1\right)=0$$
$$\iff$$
$$2\left(x+1\right)-a\left(y\right)+\left(a^{2}-2\right)\left(z+1\right)=0$$
$$-2\left(x-1\right)+a\left(y-2a\right)+\left(-a^{2}+2\right)\left(z-1\right)=0$$
But after solving the equations I see that the result is trivially true and is independent of $a$.
So how to get the value of $a$?
 A: Keep in mind that 2 non-parallel coplanar lines always intersect at exactly one 1 point. Obviously these two lines can never be parallel; their direction ratios are proportional for no value of $a$. So if they are coplanar they intersect and if they intersect they are coplanar:
$L_1,L_2$ intersect <==> $L_1,L_2$ are coplanar
The problem simplifies to asking if they intersect or not.
Let's say this intersection occurs at $t$=$t_1$ for $L_1$ and $t$=$t_2$ for $L_2$
$$(−1,0,−1)+t_1(a_0,2,0) = (1,2a_0,1)+t_2(2,3,2)$$
The system consists of 3 linear equations:$$at_1-1=1+2t_2$$$$2t_1=2a+3t_2$$$$-1=1+2t_2$$
We need to find the set of real values of $a$ for which the above system has exactly 1 solution for $(t_1,t_2)$:
Eliminating $t_2$ from the 3rd equation we need to find those values of $a$ for which the resulting system has exactly 1 solution for $t_1$(as the value of $t_2$ is 2 anyways if a value for $t_1$ exists):$$at_1=4$$$$2t_1=2a+3$$
When $a=0$, there exist no solutions for this system. Eliminating $t_1$ from eqn 1 for nonzero $a$, we get a quadratic that $a$ must satisfy: $$2a^2+3a-8=0$$This quadratic gives 2 values of a for which the lines are coplanar, the values of $t_1,t_2$ being 4/a and 1 respectively.
P.S If by "But after solving the equations" you mean you took both those planes as identical, your method is correct; either my calculation or yours is wrong.
A: Lines $L_1 = \{P+t\vec{u} : t \in \Bbb{R}\}$ and $L_2 = \{Q+t\vec{v} : t \in \Bbb{R}\}$ are coplanar if and only if $$\det(\vec u, \vec v, \overrightarrow{PQ})=0.$$
We have $\vec{PQ} = (1,2a,1)-(-1,0,-1) = (2,2a,2)$ so this determinant is
$$\det(\vec u, \vec v, \overrightarrow{PQ}) = \begin{vmatrix} a & 2 & 0 \\ 2 & 3 & 2 \\ 2 & 2a & 2 \end{vmatrix} = 2a(3-2a)$$
which is zero if and only if $a \in \left\{0,\frac32\right\}$.
When $a = 0$ then both lines are in the plane $x-z=0$, and when $a = \frac32$ then they are in the plane $8x-6y+z+9=0$.
