Prove that a given ideal is not maximal in $\mathbb C[x,y,z]$ I'm trying to prove this ideal:

$$(x^2+y^2+z^2+x+y+z,\ x^5+y^5+z^5+2(x+y+z),\ x^7+y^7+z^7+3(x+y+z))\subset
 \mathbb C[x,y,z]$$

can't be maximal.
In order to do so, I'm using the Nullstellensatz theorem and showing this ideal is not of this form: $(x-a_1,y-a_2,z-a_3)$, where $a_i\in \mathbb C$.
The problem I don't how to do this.
 A: Let $J=(x,y,z)$. We have $I+(x^2,y^2,z^2)=(x+y+z,x^2,y^2,z^2)\subsetneq J$ and then $I$ is not maximal because clearly $I\subseteq J$.
A: To show your ideal is not maximal, I'll show that the set of common zeros of your ideal contains more than one point. Taking advantage of the symmetry of your ideal, we will look for solutions that even satisfy the extra condition $x+y+z=0.$ Then (using Newton's Identities and remember $e_1=x+y+z=0$) the first equation is $$x^2+y^2+z^2+x+y+z = p_2 + e_1 = p_2 = e_1^2 - 2e_2.$$
The next one is  $$ x^5+y^5+z^5 + 2(x+y+z) = p_5 + 2e_1 = -5e_3e_2 $$ and the third is 
$$ x^7+y^7+z^7 + 3(x+y+z) = p_7 + 3e_1 = c e_3 e_2^2$$ where $c$ is a non-zero constant. 
This system is satisfied if we further impose $e_2=0.$ Hence, any common solutions of $e_1=x+y+z=0$ and $e_2 = xy+yz+xz=0$ are in the set of common zeros of your ideal. It is easy to find solutions to this that aren't the trivial $(0,0,0)$ solution. For example, pick $x=1,$ so $y+z=-1$ and $y+yz+z =0,$ which easily solves for $y= \dfrac{-1}{2} - i \dfrac{\sqrt{3}}{2} \ , \ z = \dfrac{-1}{2} + i \dfrac{\sqrt{3}}{2}.$
A: Perhaps this works?
For any $f(x,y,z) \in \mathbb{C}[x,y,z]$, it seems clear that
$$f(x,y,z)(x+y+z) \not\equiv x (\bmod \langle x^2,y^2,z^2 \rangle).$$
Therefore,
$$x \notin \langle x^2,y^2,z^2,x+y+z \rangle.$$
So we have
$$\langle x,y,z \rangle \supsetneq \langle x^2, y^2, z^2, x+y+z \rangle \supseteq I,$$
where $I$ is the ideal given by the problem statement.
