Let be the matrix the following:
$ \begin{pmatrix} P & 1 & 1 & -2 \\ 1 & P & 1 & 1 \\ 1 & 1 & P & 1 \end{pmatrix}, P \in \mathbb{R}$
I have checked the solutions for:
$ P = 1 $ there is no solution
$ P = 2 $ the solution is $ x_{1}= -2, x_{2} = 1, x_{3}= 1$
$ P = 3 $ the solution is $ x_{1}= -1, x_{2} = 1/2, x_{3}= 1/2$
I guess that for
$ P = n $ the solution is $ x_{1}= -2/(n-1), x_{2} = 1/(n-1), x_{3}= 1/(n-1)$
$ P = 0 $ the solution is $ x_{1}= 2, x_{2} = -1, x_{3}= -1$
$ P = -1 $ the solution is $ x_{1}= -1, x_{2} = 1/2, x_{3}= 1/2$
$ P = -2 $ the solution is $ x_{1}= -1/2, x_{2} = 1/4, x_{3}= 1/4$
In this case the general solution:
$ P = -n $ the solution is $ x_{-n}= -2/(n+1), x_{2} = -1/(n+1), x_{3}= -1/(n+1)$
I guess that we need to do two proofs by induction, but we are learning it right now and I have problems doing it. Thank you for you help.
$P$ is real, induction should not be the right tool. Also, when you suggest induction as the way to approach, you should check if the proposed solution indeed work for those examples that you tried.
When $P<0$, we can't have $x_1 < 0, x_2 > 0, x_3>0$ as the first entry would give us $Px_1+x_2+x_3>0$.
You should try Gaussian Elimination. That would also enable you to address the case when does it not have a solution, when does it have a unique solution, and when does it have infinitely many solutions. I would leave that to you as an exercise.
Also, you are right that when $P=1$, is has no solution.
However, let's see how I can use part of your conjecture. Your conjecture suggests that $x_2=x_3$ and $x_1=-2x_2$ and let's check:
$$\begin{bmatrix} P & 1 & 1 \\ 1 & P & 1 \\ 1 & 1 & P \end{bmatrix}\begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix}=\begin{bmatrix} -2P+2 \\ P-1 \\ P-1\end{bmatrix}=(P-1) \begin{bmatrix} -2 \\ 1 \\ 1\end{bmatrix}$$
Hence when $P \ne 1$, we indeed have
$$\begin{bmatrix} P & 1 & 1 \\ 1 & P & 1 \\ 1 & 1 & P \end{bmatrix}\begin{bmatrix} \frac{-2}{P-1} \\ \frac1{P-1} \\ \frac1{P-1} \end{bmatrix}= \begin{bmatrix} -2 \\ 1 \\ 1\end{bmatrix}$$
Hence if $P \ne 1$ and if the solution is unique, then you are done.
Let's analyze when is the matrix singular.
The determinant is a cubic polynomial in terms of $P$. When $P=1$, the determinant must be $0$, indeed all the rows are equal and the nullity is $2$. $1$ is a double root. Also, we can easily see that each row sum to $P+2$, hence $P=-2$ is another root. When $P=-2$, it is consistent and it has infinitely many solution.
When $P=-2$, the rank of the matrix is $2$ and the nullity is $1$.
$\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ is a particular solution and each row would sum to $0$.
Hence, $$\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} + t \begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}$$ is the general solution when $P=-2$.
I think it is important to be able to solve such problem using a standard way. Do solve it using Gaussian Elimination or some other ways and then compare the solution.