# Intersection point of three planes

This task comes from my text book:

For which values of parameter $$p$$ the planes intersect at one point? If such $$p$$ exists, find components of the intersection point.

$$\begin{cases} \pi_{1}:x + py - p = 0 \\ \pi_{2} : x + y - pz + p^2 + 2p - 1 = 0 \\ \pi_{3} : 2x - pz + p = 0 \end{cases}$$

Answer: For all $$p \neq -1, 0$$; the point: $$P(p^2,1-p,2p+1)$$.

Initially I thought the task is clearly wrong because two planes in $$\mathbb{R}^3$$ can never intersect at one point, because two planes are either: overlapping, disjoint or intersecting at a line.

But here I am dealing with three planes, so I think I need to find the "common intersection point".

Any tips on how to solve this task? I feel like I need to solve some kind of system of equations.

• Three planes often intersects at one point. Imagine your room: the floor, one wall, one adiacent wall – Raffaele Jun 15 at 10:23

From linear algebra, the system \begin{align*} x + py &= p \tag{1}\\ x + y - pz &= 1-2p-p^2 \tag{2}\\ 2x - pz &= -p\tag{3} \end{align*} has a unique solution if and only if $$0\neq\det\begin{pmatrix} 1&p&0\\1&1&-p\\2&0&-p \end{pmatrix}=-2p(p+1)$$ i.e., $$p\neq 0,-1$$, and in such cases the intersection point is given by $$\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix} 1&p&0\\1&1&-p\\2&0&-p \end{pmatrix}^{-1} \begin{pmatrix}p\\1-2p-p^2\\-p\end{pmatrix} =\begin{pmatrix}p^2\\1-p\\2p+1\end{pmatrix}.$$ since $$\operatorname{adj}\begin{pmatrix} 1&p&0\\1&1&-p\\2&0&-p \end{pmatrix}=\begin{pmatrix} -p&p^2&-p^2\\-p&-p&p\\-2&2p&1-p \end{pmatrix}$$
From (1), $$x=p-py$$, so \begin{align*} x &= p-py \tag{1'}\\ (1-p) y - pz &= 1-3p-p^2 \tag{2'}\\ 2py + pz &= 3p\tag{3'} \end{align*} If $$p=0$$ (3') tells us nothing and we have intersection of two planes (1') and (2') which won't give us unique solution. So assuming $$p\neq 0$$, we can divide (3') by $$p$$ and get $$z=3-2y$$. Substituting, \begin{align*} x &= p-py \tag{1'}\\ (1+p) y &= 1-p^2 \tag{2''}\\ z &= 3-2y\tag{3''} \end{align*} If $$p=-1$$, (2'') tells us nothing and again we don't have unique solution. So $$p\neq -1$$, and we have $$y=1-p, z=3-2y=2p+1, x=p-py=p^2.$$