Here is an argument that will give the existence of such good pairs, provided certain connectedness property (even a weaker property) holds, which seems reasonable, but I haven't checked the details.

Let me first try to rephrase the "goodness" condition in terms of double ratios, where I use hermitian forms instead of matrices for simplicity, i.e. write $A(v,w)$ instead of $\langle Av, w\rangle$.

Suppose a pair $(v,w)$ is good for both $A$ and $B$. Then $A(v,v)=A(w,w)$ and $B(v,v)=B(w,w)$ imply that the double ratio

$$
R(A,B;v,w):= \frac{A(v,v)}{A(w,w)} : \frac{B(v,v)}{B(w,w)} = 1.
$$

Vice versa, suppose $R(A,B;v,w)=1$. Then I can always scale one of the vectors, say $v$, to achieve $A(v,v)=A(w,w)$, which together with the double ratio relation will simultaneously imply $B(v,v)=B(w,w)$. Thus, up to scaling, the "goodness" of $(v,w)$ for both $A$ and $B$ is equivalent to their double ratio being equal to $1$, together with the orthogonality $A(v,w)=B(v,w)=0$. In view of this equivalence, we shall in sequel always mean "good up to scaling (of one of the vectors)" without explicitly mentioning it.

Now consider any pair $(v,w)$ for which $R(A,B;v,w)\ne 1$. Note that $R$ is always positive. Then switching $v$ and $w$ leads to the inverse of their double ratio:
$$
R(A,B;w,v) = R(A,B;v,w)^{-1}.
$$
Hence one of these ratios is $<1$ and the other is $>1$.

Next, let us bring up the orthogonality and consider the (real-algebraic) set $O(A,B)$ of all $(v,w)$ with $A(v,w)=B(v,w)=0$. I can arbitrary choose $v$ and then $w$ in the intersection of its both orthogonal complements with respect to $A$ and $B$. In particular, $O(A,B)$ is connected. Furthermore, for any pair $(v,w)\in O(A,B)$, either it is already good for $A,B$ or any path in $O(A,B)$ connecting $(v,w)$ with $(w,v)$ must have pairs with double ratio on both sides of $1$, hence the path must contain at least one good pair.

Fixing $v$, the set of all $w$ for which $(v,w)$ is $(A,B)$-good and $C$-orthogonal, if nonempty, is generically a 1-torus (generating a complex line bundle via rescaling), when $A,B,C$ are also generic. Here we rely on the property that generically the $(A,B,C)$-orthogonal complement of $v$ is 1-dimensional.
The whole real-alebraic variety $G(A,B;C)$ of all pairs $(v,w)$ good for $A,B$ and $C$-orthogonal is therefore a semialgebraic torus bundle (with possible singular fibers corresponding to degenerations of the orthogonal complements) over a semialgebraic subset of codimension $1$ in $\mathbb C^4$.

Furthermore, I think that $G(A,B;C)$ should be connected, due to the above strong property that it must meet every path connecting pairs $(v,w)$ and $(w,v)$. I haven't checked the details that may require some use of topology.

Now, assuming $G(A,B;C)$ is connected, we can repeat the above path argument with the double ratios $R(A,C;v,w)$ to achieve the same conclusion, i.e. every path in $G(A,B;C)$ connecting $(v,w)$ with $(w,v)$ must contain a pair good for all $A,B,C$.

Thus, to complete the arguments, it would suffice to find a single pair $(v,w)\in G(A,B;C)$ which is in the same connected component as the flip $(w,v)$ (which is weaker than the connectedness of $G(A,B;C)$).

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