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.
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 $$ \operatorname{adj}\begin{pmatrix} 1&p&0\\1&1&-p\\2&0&-p \end{pmatrix}=\begin{pmatrix} -p&p^2&-p^2\\-p&-p&p\\-2&2p&1-p \end{pmatrix} $$
But let's assume you don't know linear algebra.
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 $$ y=1-p, z=3-2y=2p+1, x=p-py=p^2. $$