Find intersection of all plan $n^{2}x+(2n-1)y+nz=3$

Question

Problem : Let $(P_{n})_{n}$ the families of plan equation :

$$P_{n}~~:~~n^{2}x+(2n-1)y+nz=3$$ And note that $E$ are the intersection of all this plan e.o :

$$E=\{~M(x,y,z)~/~\forall n\in\mathbb{N}~,M\in P_{n} \}$$

Choose the right option :

$E=?$

$•~\varnothing$

$•~E~\text{is the plan}~ x+y+z=3$

$•~E~\text{is the droit}~\begin{cases}x+y+z=3,\\y=-3\end{cases}$

$•~E~\text{is the point} (0,-3,6)$

I don't know how but for $n=1$ and $n=0$ the intersection is the suggestion 3 But for all I don't know Can you assist!

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Thanks!

Answer 1

If you solve the system you get for $n\in\{1,2,3\}$: \begin{eqnarray}x+y+z&=&3\\ 4x+3y+2z&=&3\\ 9x+5y+3z&=&3\\ \end{eqnarray} which is a single point $(0,-3,6)$. Now check if it lies in every plane: $$ n^2\cdot 0+(2n-1)\cdot (-3) +n\cdot 6 = 3$$ which is true for all $n$, so you chose last answer.