What exactly do you mean by "paramatrization"? Anyway, since your equation is homogeneous, you're really asking for $\mathbb Q$-rational points on the surface $S$ in $\mathbb P^3$ defined by your equation. The surface $S$ is an example of what is known as a K3 surface. Most likely what you have in mind for a paramatrization would be homogeneous polynomials $x(u,v,w),y(u,v,w),z(u,v,w),t(u,v,w)\in\mathbb Z[u,v,w]$ so that you get all solutions by plugging in integer values of $u$, $v$, and $w$. Such a solution very likely does not exist. Here's why. First, there is no way to parametrize the complex solutions using rational functions $x(u,v,w),y(u,v,w),z(u,v,w),t(u,v,w)\in\mathbb C[u,v,w]$, because a K3 surface is not birationally equivalent to $\mathbb P^2$. So that means there are two possibilities. First, the solutions are Zariski dense, in which case you won't have a parametrization. Second, the solutions lie on a finite number of curves. (I don't know offhand how to distinguish between these possibilities.) However, is is conjectured that for any K3 surface, there is a finite extension $K/\mathbb Q$ such that the $K$-rational points are Zariski dense. So if it happens that you can parametrize the integer solutions to your particular equation, that's in some sense because your field $\mathbb Q$ isn't big enough to really reflect the arithmetic structure of the problem.

What exactly do you mean by "paramatrization"? Anyway, since your equation is homogeneous, you're really asking for $\mathbb Q$-rational points on the surface $S$ in $\mathbb P^3$ defined by your equation. The surface $S$ is an example of what is known as a K3 surface. Most likely what you have in mind for a paramatrization would be homogeneous polynomials $x(u,v,w),y(u,v,w),z(u,v,w),t(u,v,w)\in\mathbb Z[u,v,w]$ so that you get all solutions by plugging in integer values of $u$, $v$, and $w$. Such a solution very likely does not exist. Here's why. First, there is no way to parametrize the complex solutions using polynomial functions $x(u,v,w),y(u,v,w),z(u,v,w),t(u,v,w)\in\mathbb C[u,v,w]$, because a K3 surface is not birationally equivalent to $\mathbb P^2$. So that means there are two possibilities. First, the integer solutions are Zariski dense, in which case you won't have a parametrization. Second, the integer solutions lie on a finite number of curves. (I don't know offhand how to distinguish between these two possibilities.) However, is is conjectured that for any K3 surface, there is a finite extension $K/\mathbb Q$ such that the $K$-rational points are Zariski dense. So if it happens that you can parametrize the integer solutions to your particular equation, that's in some sense because your field $\mathbb Q$ isn't big enough to really reflect the arithmetic structure of the problem. And the solutions won't be that interesting, since they'd only be a finite number of one-parameter families.