This is the problem:
In $\mathbb{E}^3$ we consider the conic $\gamma$ of equations $x=yz-2=0$ , the line $a$ of equations $x=y+z=0$ and the surface $Q$, that is generated by the rotation of $\gamma$ around $a$.
(a) After having explained the equation, show that $Q$ is a hyperbolic hyperboloid.
(b) Describe the cartesian equations of the straight lines contained in Q, and the minimum circle of Q.
(c) Let $Q'$ projective closure of $Q$ in $\mathbb{P} ^ 3$, determine a plane $\pi$ in $\mathbb{P} ^ 3$, through P = (2:0:0:1), such that the trace of Q' in the space affine $\mathbb{A}: = \mathbb{P} ^ 3/\pi$ is a paraboloid
I try to complete (a): I found the plane perpendicular to $ a $, but I do not know how to impose conditions to generate Q
Let's see about (a). $x=0$ tells you that both your curves live in the $y$-$z$-plane. So you might want to draw or plot an image of the situation in that plane.
The equation $y+z=0$ for $a$ can be re-interpreted as the line through the origin and in direction $(0,1,-1)$. One possible approach is rotating your coordinate system in such a way that this axis coincides with one of the coordinate axes. To map that line onto the $y$ axis, you can use
$$ M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac12\sqrt2 & -\frac12\sqrt2 \\ 0 & \frac12\sqrt2 & \frac12\sqrt2 \end{pmatrix} $$
A point $p=(x,y,z)$ lies on the rotated curve $\gamma'$ iff its preimage lies on $\gamma$, i.e. if $\gamma(M^{-1}\cdot p)=0$.
$$\gamma': x=\tfrac12z^2-\tfrac12y^2-2=0$$
Now you can rotate that around the $y$ axis. The term $z^2$ in the above equation describes the squared distance from the $y$ axis in the $y$-$z$-plane, so you rotate by writing $y^2+z^2$ instead of $z^2$.
$$Q': \tfrac12x^2+\tfrac12z^2-\tfrac12y^2-2=0$$
Now rotate everything back. A point $p$ lies on $Q$ if its transformed version $M\cdot p$ lies on $Q'$, i.e. $Q'(M\cdot p)=0$.
$$Q: \tfrac12x^2+yz-2=0$$
Now you have the equation of the hyperboloid. To see that it is hyperbolic, you can intersect the hyperboloid with the line. Simply insert a generic point on the line, i.e. $\lambda\cdot(0,1,-1)$, into the equation of the hyperboloid.
$$Q\left(\lambda\cdot\begin{pmatrix}0\\1\\-1\end{pmatrix}\right) = -\lambda^2-2=0$$
This equation has no real solution, which means that the hyperboloid does not intersect the line. Therefore your hyperboloid has only a single sheet. I assume that this is what you mean by “hyperbolic” in this context. If not, please clarify since I can't find a good reference of this use of the term. Note that this decision of the line not intersecting the hyperboloid could have been deduced from $\gamma$ and $a$ alone, without the equation for $Q$.