This isn't a solution, just a comment that got too long for the comment box: Assuming that all finite groups occur as Galois groups over $k$, the answer to this question should only depend on the characteristic of $k$.

Consider the more detailed question:

For a finite group $G$, and subgroups $H_1$, $H_2$ and $H_3$, is there a Galois extension $L$ of $k$ with Galois group $G$, and elements $v_i > \in L$ such that the stabilizer of $v_i$ is $H_i$ and $v_1+v_2+v_3=0$.

I claim that, for given $(G, H_1, H_2, H_3)$, assuming that there is some $G$-extension of $k$, the answer to this question only depends on the characteristic of $k$. Proof: As a $G$-representation, $L$ is the permutation representation on $X:=G/(H_1 \cap H_2 \cap H_3)$. The question, then, is whether we can find $L$-valued functions, $f_i$ on $X$, such that $f_i$ is constant on $H_i$ orbits (but not for any larger subgroup) and $f_1+f_2+f_3=0$. This is a collection of linear equalities and inequalities in $3 |X|$ variables, with integer coefficients. So whether or not they have a solution depends only on the characteristic of $k$. (To see that the characteristic can matter, take $G=S_3$, $H_1$ and $H_2$ two different subgroups of order $2$ and $H_3$ the subgroup of order $3$. You should get a solution in characteristic $3$, and not otherwise.)

Of course, answering the original question just means answering this question for all $(G, H_1, H_2, H_3)$ with $|G/H_i| = d_i$. (One can immediately make two reductions. First, a necessary condition is that $H_1 \cap H_2 = H_1 \cap H_3= H_2 \cap H_3$. Second, one can immediately reduce to the case that $H_1 \cap H_2$ contains no nontrivial normal subgroup. The latter means that $|G| \leq (d_1 d_2)!$, so the problem is finite.)

I see no reason to believe that you will get a nicer answer by forgetting the groups and only remembering the degrees, but of course I haven't thought very hard about the problem.

This isn't a solution, just a comment that got too long for the comment box: Assuming that all finite groups occur as Galois groups over $k$, the answer to this question should only depend on the characteristic of $k$.

Consider the more detailed question:

For a finite group $G$, and subgroups $H_1$, $H_2$ and $H_3$, is there a Galois extension $L$ of $k$ with Galois group $G$, and elements $v_i > \in L$ such that the stabilizer of $v_i$ is $H_i$ and $v_1+v_2+v_3=0$.

I claim that, for given $(G, H_1, H_2, H_3)$, assuming that there is some $G$-extension of $k$, the answer to this question only depends on the characteristic of $k$. Proof: As a $G$-representation, $L$ is the permutation representation on $X:=G/(H_1 \cap H_2 \cap H_3)$. The question, then, is whether we can find $L$-valued functions, $f_i$ on $X$, such that $f_i$ is constant on $H_i$ orbits (but not for any larger subgroup) and $f_1+f_2+f_3=0$. This is a collection of linear equalities and inequalities in $3 |X|$ variables, with integer coefficients. So whether or not they have a solution depends only on the characteristic of $k$ (we are using that $k$ is infinite). To see that the characteristic can matter, take $G=S_3$, $H_1$ and $H_2$ two different subgroups of order $2$ and $H_3$ the subgroup of order $3$. You should get a solution in characteristic $3$, and not otherwise.

Of course, answering the original question just means answering this question for all $(G, H_1, H_2, H_3)$ with $|G/H_i| = d_i$. (One can immediately make two reductions. First, a necessary condition is that $H_1 \cap H_2 = H_1 \cap H_3= H_2 \cap H_3$. Second, one can immediately reduce to the case that $H_1 \cap H_2$ contains no nontrivial normal subgroup. The latter means that $|G| \leq (d_1 d_2)!$, so the problem is finite.)

I see no reason to believe that you will get a nicer answer by forgetting the groups and only remembering the degrees, but of course I haven't thought very hard about the problem.

This isn't a solution, just a comment that got too long for the comment box: Assuming that all finite groups occur as Galois groups over $k$, the answer to this question should only depend on the characteristic of $k$.

Consider the more detailed question:

For a finite group $G$, and subgroups $H_1$, $H_2$ and $H_3$, is there a Galois extension $L$ of $k$ with Galois group $G$, and elements $v_i > \in L$ such that the stabilizer of $v_i$ is $H_i$ and $v_1+v_2+v_3=0$.

I claim that, for given $(G, H_1, H_2, H_3)$, assuming that there is some $G$-extension of $k$, the answer to this question only depends on the characteristic of $k$. Proof: As a $G$-representation, $L$ is the permutation representation on $X:=G/(H_1 \cap H_2 \cap H_3)$. The question, then, is whether we can find $L$-valued functions, $f_i$ on $X$, such that $f_i$ is constant on $H_i$ orbits (but not for any larger subgroup) and $f_1+f_2+f_3=0$. This is a collection of linear equalities and inequalities in $3 |X|$ variables, with integer coefficients. So whether or not they have a solution depends only on the characteristic of $k$. (To see that the characteristic can matter, take $G=S_3$, $H_1$ and $H_2$ two different subgroups of order $2$ and $H$ the subgroup of order $3$. You should get a solution in characteristic $3$, and not otherwise.)

Of course, answering the original question just means answering this question for all $(G, H_1, H_2, H_3)$ with $|G/H_i| = d_i$. (One can immediately make two reductions. First, a necessary condition is that $H_1 \cap H_2 = H_1 \cap H_3= H_2 \cap H_3$. Second, one can immediately reduce to the case that $H_1 \cap H_2$ contains no nontrivial normal subgroup. The latter means that $|G| \leq (d_1 d_2)!$, so the problem is finite.)

I see no reason to believe that you will get a nicer answer by forgetting the groups and only remembering the degrees, but of course I haven't thought very hard about the problem.

This isn't a solution, just a comment that got too long for the comment box: Assuming that all finite groups occur as Galois groups over $k$, the answer to this question should only depend on the characteristic of $k$.

Consider the more detailed question:

For a finite group $G$, and subgroups $H_1$, $H_2$ and $H_3$, is there a Galois extension $L$ of $k$ with Galois group $G$, and elements $v_i > \in L$ such that the stabilizer of $v_i$ is $H_i$ and $v_1+v_2+v_3=0$.

I claim that, for given $(G, H_1, H_2, H_3)$, assuming that there is some $G$-extension of $k$, the answer to this question only depends on the characteristic of $k$. Proof: As a $G$-representation, $L$ is the permutation representation on $X:=G/(H_1 \cap H_2 \cap H_3)$. The question, then, is whether we can find $L$-valued functions, $f_i$ on $X$, such that $f_i$ is constant on $H_i$ orbits (but not for any larger subgroup) and $f_1+f_2+f_3=0$. This is a collection of linear equalities and inequalities in $3 |X|$ variables, with integer coefficients. So whether or not they have a solution depends only on the characteristic of $k$. (To see that the characteristic can matter, take $G=S_3$, $H_1$ and $H_2$ two different subgroups of order $2$ and $H_3$ the subgroup of order $3$. You should get a solution in characteristic $3$, and not otherwise.)

Of course, answering the original question just means answering this question for all $(G, H_1, H_2, H_3)$ with $|G/H_i| = d_i$. (One can immediately make two reductions. First, a necessary condition is that $H_1 \cap H_2 = H_1 \cap H_3= H_2 \cap H_3$. Second, one can immediately reduce to the case that $H_1 \cap H_2$ contains no nontrivial normal subgroup. The latter means that $|G| \leq (d_1 d_2)!$, so the problem is finite.)

I see no reason to believe that you will get a nicer answer by forgetting the groups and only remembering the degrees, but of course I haven't thought very hard about the problem.