Why is family of planes not working here?

Question

Came across a question where it said $L_1$ 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 $x+2y−z−3=0$, $3x−y+2z−1=0$.

I understand the basic approach where we can find $L_1$ and $L_2$ individually, and the plane containing them comes out to be $7x-7y+8z+3=0$.

I note that this plane should be of the form $2x-2y+3z−2+ A(x−y+z+1)=0$, and also of the form $x+2y−z−3 +B(3x−y+2z−1)=0$, for arbitrary constants $A$ and $B$.

Now $A=5$ seems to work, but no value of $B$ fits. My query is, why is that the case?

Answer 1

The problem is that, if $P(x, y, z) = 0$ is the equation of a plane (or indeed any equation), then $\lambda P(x, y, z) = 0$ represents the same plane (or whatever other object) so long as $\lambda \neq 0$. In this case, try $B = -\frac{3}{2}$, and you'll see that you end up with the final answer, but scaled by $-\frac{1}{2}$.

Answer 2

I don't see any problem with your line of thinking. It is just that a bit of rigour is required.

Following from where you left of we have by comparing the coefficients of $x,y,z$ from the family of planes you obtained we get $$\frac{2+A}{1+3B}=\frac{-(2+A)}{2-B}=\frac{3+A}{-1+2B}=\frac{A-2}{-3-B}$$ Solving for $A$ and $B$

we get $A=5$ and $B=-\frac{3}{2}$

which matches with the equation of plane you obtained