Parametrization of integral solutions of $3x^2+3y^2+z^2=t^2$ and rational solutions of $3a^2+3b^2-c^2=-1$ 1/ Is it known the parameterisation over $\mathbb{Q}^3$ of the solutions of
$3a^2+3b^2-c^2=-1$
2/   Is it known the parameterisation over $\mathbb{Z}^4$ of the solutions of
$3x^2+3y^2+z^2=t^2$
References, articles or books are welcome
Sincerely, John
 A: For #1, we can take a particular solution such as $(a_0,b_0,c_0)=(0,0,1)$, and search parametric solution in the form: $(a,b,c)=(a_0+\alpha t, b_0+\beta t, c_0+t)$. Plugging it into the equation and solving for $t\ne 0$, we get:
$$t = \frac{2}{3\alpha^2 + 3\beta^2 - 1}.$$
So, we get rational parametrization with parameters $\alpha,\beta\in\mathbb Q$:
$$(a,b,c) = \bigg(\frac{2\alpha}{3\alpha^2 + 3\beta^2 - 1},\ \frac{2\beta}{3\alpha^2 + 3\beta^2 - 1},\ 1 + \frac{2}{3\alpha^2 + 3\beta^2 - 1}\bigg).$$

For #2, we can similarly parametrize $3\left(\frac{x}{t}\right)^2 + 3\left(\frac{y}{t}\right)^2 + \left(\frac{z}{t}\right)^2 = 1$ and then explicitly expand parameters as fractions, and set $t$ be the common denominator. This way we get parametrization:
$$(x,y,z,t) = \frac{p}{q}\bigg(-2uw,\ -2vw,\ 3u^2 + 3v^2 - w^2,\ 3u^2 + 3v^2 + w^2\bigg),$$
where parameters $u,v,w\in\mathbb Z$, and parameters $p,q\in\mathbb Z$ allow to scale the variables, with the requirement that $q$ represents a common divisor of the variables.
A: on the second question $3x^2 + 3 y^2 + z^2 = t^2$  in integers:
One description is this:  if $x,y$  are integers and not both odd, then $3 (x^2 + y^2)  \neq 2 \pmod 4.$ As a result it may be   expressed a few ways as $t^2 - z^2 = (t+z)(t-z)$ When odd, we may take $t-z$   to be any factor of  $3 (x^2 + y^2)$  because $t+z$ will be the same $\pmod 2.$  When  $ N =3 (x^2 + y^2)$  is divisible by $4,$ we take only divisors $d$  where both $d$ and $N/d$  are even. Further conditions make the quadruple primitive.
As far as integer parametrization,   it turned out that two evident recipes were enough.  Those (primitive) quadruples with $x,y$ even appear as ( we need $j+k+l+m$ odd )
$$ x = 2(jl -  km) \; , \; \; y = 2 (jm +kl) $$
$$  z = 3 j^2 + 3 k^2 - l^2 - m^2 \; , \; \;  t = 3 j^2 + 3 k^2 + l^2 + m^2  $$
Those with $x+y$  odd appear as
$$  x = j^2 + k^2 - l^2 - m^2 \; , \; \; y = 2 (jl-km)  $$
$$ z = 4(jm+kl) + (j^2 + k^2 + l^2 + m^2 ) $$
$$ t = 2(jm+kl) + 2(j^2 + k^2 + l^2 + m^2 ) $$
The closest comparison I have is for ternary quadratic forms:  Mordell,  in Diophantine Equations,  page 47, Theorem 4, says the the isotropic vectors of an integer ternary come as a finite set of recipes, each one of type $x = a_1 p^2 + b_1 pq + c_1 q^2 \; , \; \; \;  $ $y = a_2 p^2 + b_2 pq + c_2 q^2 \; , \; \; \;  $ $z = a_3 p^2 + b_3 pq + c_3 q^2 \; , \; \; \;  $
The interesting bit is that quaternary forms need recipes with four variables. A famous example is Pythagorean Quadruples ,   $x^2 + y^2 + z^2 = m^2$.  There is a nice writeup by Robert Spira in 1962;   he says the first correct proof (that all quadruples come from the given parametrization )   was Dickson in 1920.
