How to find the distance from origin of a plane containing given lines $L_1$ and $L_2$?

Question

If $L_1$ is the line of intersection of the planes $2x-2y+3z-2=0$,$x-y+z+1=0$ and $L_2$ is the line of intersection of the planes $x+2y-z-3=0$,$3x-y+2z-1=0$, then the distance of the origin from the plane, containing the lines $L_1$ and $L_2$ is:-

I have tried it this way:

The equation of any plane passing through $L_1$: $2x-2y+3z-2+a(x-y+z+1)=0$ (where $a$ is an arbitrary constant)

The equation of any plane passing through $L_2$: $x+2y-z-3+b(3x-y+2z-1)=0$ (where $b$ is an arbitrary constant)

Now, the above two equations should represent the same plane so the coefficients will be in the same ratio. After solving for $a$ and $b$ we will get more than one value for each of them; I don't know what am I am doing wrong to obtain the equation of the plane. I think I might be getting the equation of the lines of intersection of $p_1$ with $p_3$ and $p_4$ and so on. Please help!

Answer 1

What is said at the beginning is false: let's take a simple example to convince ourselves of this. Let $L$ the line of intersection of the planes $y=0$ and $x=0$; then $x=0$ is not of the form $y+ax=0$, whatever the value of $a$ chosen.

That being said, here's how you can do it :

1. Let's write $L_1$ in a way that is easier to handle. $M=(x,y,z)\in L_1\iff\begin{cases}2x-2y+3z=2\\x-y+z=-1\end{cases}\iff\begin{cases}z=4\\x-y+z=-1\end{cases}\iff\begin{cases}z=4\\x=y-5\end{cases}\iff (x,y,z)=(-5,0,4)+y(1,1,0)\iff M\in(-5,0,4)+\mathbb R(1,1,,0)$
2. In the same way, $M\in L_2\iff M \in (\frac57,\frac87,,0)+\mathbb R(-3,5,7)$
3. You can verify that $J=(-1,4,4)\in L_1\cap L_2$
4. Let $v_1=(1,1,0), v_2=(-3,5,7). n=v_1\times v_2=(7,-7,8), \|n\|=9\sqrt2$. Then the distance from the origin to the plane is given by $d=|\vec{OM}.n|$, for any $M$ in the plane; for example, with $M=J, d=\frac{1}{9\sqrt2}|(-1,4,4).(7,-7,8)|=\frac{\sqrt2}{6}$