Let $\Pi_1 : 7x-5y-2z=0, \Pi_2 : 5x-4y-z=0$ and $\mathbb{L}$ the line that passes through points $P=(-2,3,-3)$ and $Q=(-1,2,-1)$. Find all planes $\Pi$ that satisfy:
- $\Pi \cap \Pi_1 \cap \Pi_2 = \emptyset$
- $d(R,\Pi)=\sqrt{14} \forall R \in \mathbb{L}$
So far, I've found out that $\Pi_1\cap\Pi_2= { X : \lambda (1,1,1) } $ and $\mathbb{L} : X=\lambda(1,-1,2)+(-1,2,-1)$; given $\Pi : NX = NQ$, with $N$ being an orthogonal vector to the plane and $Q$ being any point in $\Pi$, it follows that $N \perp(1,1,1)$ and $N\perp(1,-1,2)$, so we can define $N=(1,1,1)\times(1,-1,2)=(3,-1,-2)$.
As $d(R,\Pi) = \frac{|N(R-Q)|}{||N||} = \sqrt{14}$
$\frac{|(3,-1,-2)((-3\lambda-1,\lambda+2,-4\lambda-1)-(x,y,z))|}{||(3,-1,-2)||} = \sqrt{14}$
$|(3,-1,-2)(-3\lambda-1-x,\lambda+2-y,-4\lambda-1-z)| = 1$
$-9\lambda-3-3x -\lambda-2+y +8\lambda+2+2z = \pm 1$
$-3x+y+2z-2\lambda-3 = \pm 1$
and I simply don't know what to do next. Any help would be really appreciated.
EDIT: Solved.
$d(R,\Pi) = \frac{|N(R-Q)|}{||N||} = \sqrt{14}$
$(3,-1,-2)((-1,2,1)-(x,y,z)) = \pm 1$
$-3x+y+2z-7=\pm 1$
So we can take $Q$ as either $(0,4,2)$ or $(0,2,2)$ and we get two different planes:
$\Pi_3: 3x-y-2z=-8$ and $\Pi_4: 3x-y-2z=-6$. Once again, thanks a lot.