Solve the system for unknowns $x,y,z$ and parameters $a,b,c$ \begin{cases} {x}^{2}+{y}^{2}+{z}^{2}={a}^{2}+{b}^{2}+{c}^{2},\\ ax+by+cz+a^2+b^2+c^2=0. \end{cases}
We have two equations and three unknowns. I can solve it for two unknowns only. Any ideas?
Rewrite the first equation as: $$a^2+b^2+c^2=\frac{x^2+y^2+z^2+a^2+b^2+c^2}{2}$$ Substitute in the second equation, then double it, and you get:
$$(a+x)^2+(b+y)^2+(c+z)^2=0$$
The algebraic explanation of Thomas Andrews is very straightforward.
Nevertheless, it can be interesting to have a geometric interpretation of this issue.
Let $V=\pmatrix{a\\b\\c}$ and $v=\|V\|=\sqrt{a^2+b^2+c^2}$
The first equation represents a sphere centered in the origin with radius $v$.
The 2nd equation represents a plane which is at distance
$$\dfrac{a0+b0+c0+(a^2+b^2+c^2)}{\sqrt{a^2+b^2+c^2}}=\sqrt{a^2+b^2+c^2}=v$$
from the origin (see formula for [Distance from a point to a plane] (https://en.wikipedia.org/wiki/Plane_(geometry))).
Thus it is a tangent plane to the sphere.
Therefore there is a unique point of contact which is clearly
$$(x,y,z)=(-a,-b,-c)$$
By Cauchy-Schwartz $$a^2+b^2+c^2\overset{(2)}{=}-(ax+by+cz)\overset{C.S.}{\leq} (a^2+b^2+c^2)^{1/2}(x^2+y^2+z^2)^{1/2}\overset{(1)}{=}a^2+b^2+c^2$$ therefore they are all equalities. Now, the equality in C.S. holds iff $(a,b,c)=\lambda (x,y,z)$ with $\lambda\in\mathbb R$. Substituting in the second equation you get $\lambda+\lambda^2=0$, which has solutions $0$ and $-1$. If $\lambda=0$ then $a=b=c=x=y=z=0$, otherwise $a=-x,b=-y,c=-z$ (which actually includes the former).