Let us describe the general case first, before going to your special case. The intersection of two quadrics has always degree $4$, if you count the intersection with multiplicities. So if you want something of degree $3$, it means that it is a conic (degree $2$) and a line counted with multiplicity $2$, or three lines where one is counted with multiplicity $2$.

If you want to find how is the intersection, you can choose one quadric to be smooth, and then isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ over $\mathbb{C}$, and restricts the equation of the second one to the first one. You get an equation of bidegree $2,2$ and check if it decomposes. If the quadric is a cone you can do a similar argument.

But in general, if you have the explicit equation it is not hard to directly check if some line is contained in both quadrics.

Let us go back to your equation. It is quite strange written, since you let coefficients which are zero, but it corresponds simply to two equation of the form

$q_2(Z_1Z_3-Z_2Z_4)+q_3(Z_2Z_3+Z_1Z_4)+q_6Z_3Z_4+q_7(Z_3^2-Z_4^2)+q_8(Z_3^2+Z_4^2)$

Your other assumptions on the $q_i$'s do not seem to be really special. If you have no conditions on the $a_i,l_i$, the $q_i$ above can be anything.

As Johannes observed, the line $Z_3=Z_4=0$ is contained in your quadrics. This implies that the intersection contains the line (counted maybe with multiplicity) and the remains curve has degree $3$. If you check correctly you will see that there is no reason that in fact the intersection has degree $3$ in your case; it is easy to find coefficients such that the intersection is the line $Z_3=Z_4$ and an irreducible cubic.

If someone told you that the intersection had degree $3$, he probably meant that the intersection is the line AND a curve of degree $3$.