Do anyone know any universal and simple method do determine the number of solution (points in $\mathbb{A}^{3}$ affine space)in a quadric over a finite field?
For example the number of points in the following quadric \begin{equation} - 3 x_{1}^{2} + 6x_{1}x_{2} + x_{1}x_{3} - 6x_{1} - 3x_{2}^{2} − x_{2}x_{3} + 6x_{2} + 7x_{3}^{2} + x_{3} − 2 = 0 \end{equation} over a finite field $\mathbb{F}_{17} = \mathbb{Z}/17\mathbb{Z}$?
How does it work if the space is $\mathbb{A}^{2}$?
On
HINT:
To find the (number of) zeroes of $F(x_1, x_2, x_3) $, a polynomial of degree $d$ , consider the homogenized problem $y_0^d \cdot F(\frac{y_1}{y_0}, \frac{y_2}{y_0}, \frac{y_3}{y_0}) = 0$. Now we have to look at solutions for $$G(y_0, y_1, y_2, y_3) = 0$$ the advantage being that $G$ is homogenous ( of degree $d$). To get back to your problem, out of the homogenous (projective) solutions for $G(y_0, y_1, y_2, y_3)=0$ you substract the (projective) solutions where $y_0=0$, that is, solutions for $G(0, y_1, y_2, y_3)=0$.
So now the problem is counting the number of (projective) solutions for $G(y_0, y_1, y_2, y_3)=0$. Since there are $(q-1)$ affine (non-zero ) points going to a projective point $[y_0\colon y_1\colon y_2 \colon y_3]$, the number of projective solutions for a homogenous equation equals $\frac{1}{q-1} ( \textrm{ # affine solutions} -1)$. (the numerator is $\# \ -1$ since we eliminate the affine solution $(0,0,\ldots,0)$)
Now on this site it is treated the problem of finding the number of solutions for a homogenous quadratic polynomial over a finite field with $q$ element $\mathbb{F}_q$ ($q$ odd).
You now have some tools towards the solution.
I will give the answer based on the posted sample quadric. (Because somebody constructed some special joke with it, then inoffensively stated the joke as a trap / as a problem.)
Instead of $x_1$, $x_2$, $x_3$ i will use $x,y,z$ for an easy typing. It is convenient to get the homogenized version of the equation, so we add a new variable $x_0=w$ for this, and introduce: $$ \begin{aligned} F(x,y,z,w) &:= -3x^2 + 6xy + xz - 6xw - 3y^2 − yz + 6yw + 7z^2 + zw − 2w^2 \\ &= w^2 - 3(x-y-3z+w)^2 \ . \\[3mm] & \qquad\text{The dehomogenized version is the given one:} \\[3mm] f(x,y,z) &:=F(x,y,z,1) \\ &= 1 - 3(x-y-3z+1)^2 \ . \end{aligned} $$ Since $3$ is not a quadratic residue modulo $17$, there are no affine points $(x,y,z)\in \Bbb A^3(\Bbb F_{17})=\Bbb F_{17}^3$ with $f(x,y,z)=0$.
Further comments: The affine space is maybe not the right place to have a "structural counting". If we are searching for points in the projective space, $P=[x:y:z:w]\in \Bbb P^3(F_{17})$ with $F(x,y,z,w)=0$, then we get some. For the same reason, $3$ not a quadratic residue, the point $P$ must have $w=0$ and $(x-y-3z+w)=0$, the intersection of two hyperplanes. Explicitly we count points $P\in \Bbb P^2(\Bbb F_{17})\subset \Bbb P^3(\Bbb F_{17})$ $$ P=[x:y:z]\to[x:y:z:0]\in \Bbb P^3(\Bbb F_{17}) $$ that satisfy $t:=(x-y-3z)=0$. Then $t=0$ is a hyperplane $\cong \Bbb P^1$ inside $\Bbb P^2$. It has $17+1$ elements.
In the general case, if there is a rational point $P$ of a given quadric $C$ defined over some finite ground field $F$, then consider the lines through $P$ in the given (affine or projective) space. In the generic case such a line intersects $C$ in exactly one further point. Since we can count the number of lines through $P$, and the special (non-generic) cases, we get the number of points on $C$. This is what happens in the cited MO-link from the post. Well, we can apply this counting strategy if we have an $F$-rational point on $C$. This is not the case for the given affine quadric. It was constructed to not have such a point.