 $\alpha+\beta$ be?
I believe the answer is as follows : the degree of $\alpha+\beta$ can equal some
value $d$ iff
(1) $d \leq ab$ and $a \leq db$ and $b \leq da$. (this condition is obviously
necessary)
(2) $d$ divides $ab$, or $a$ divides $db$, or $b$ divides $da$.
The "if" part probably involves Galois theory as in Gerry's answer to the special case.  
EDIT 07/01/2010 : As Gerry noted, the conjecture above is grossly false. Below is a "corrected version"
of my conjecture.
I believe the answer is as follows : the degree of $\alpha+\beta$ can equal some
value $d$ iff
(*)  There is some integer $e$ divisible by all of $a,b,d$, and lower than or equal to
all of $ab,ad,bd$ (this is a necessary condition, as is seen by taking $e$ to be the degree
of the extension $k(\alpha,\beta)/k$, where $k$ is the base field). 

A:

I'm not sure what the answer is, but it can't be what you suggest. Your conditions allow $a=3$, $b=6$, 
$d=7$. But ${\bf Q}(\alpha+\beta)$ is contained in $K={\bf Q}(\alpha,\beta)$, so $d$ must divide the degree of $K$, which must be a multiple of 6 (since it contains ${\bf Q}(\beta)$) but no greater than 18. 

A:

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
$ \alpha+\beta$ be? 
 I believe the answer is as follows: the degree of $ \alpha+\beta$ can equal some 
 value $ d$ iff 
 (1) $ vitamin d \leq ab$ and $ a \leq db$ and $ b \leq da$. (this circumstance is obviously 
 necessary) 
 (2) $ d$ divides $ ab$, or $ a$ divides $ db$, or $ b$ separate $ da$. 
 The " if " function probably involves Galois hypothesis as in Gerry's answer to the special case.  
 EDIT 07/01/2010: As Gerry note, the conjecture above is grossly false. Below is a " compensate version " 
 of my conjecture. 
 I think the answer is as follows: the degree of $ \alpha+\beta$ can equal some 
 value $ d$ iff 
 (*)   There is some integer $ e$ divisible by all of $ a, b, d$, and lower than or equal to 
 all of $ ab, ad, bd$ (this is a necessary circumstance, as is seen by taking $ e$ to be the degree 
 of the reference $ k(\alpha,\beta)/k$, where $ k$ is the base field). 

 ampere: 

 I'm not sure what the solution is, but it can't be what you suggest. Your condition allow $ a=3 $, $ b=6 $, 
 $ d=7$. But $ { \bf Q}(\alpha+\beta)$ is contained in $ K={\bf Q}(\alpha,\beta)$, so $ d$ must separate the degree of $ K$, which must be a multiple of 6 (since it contains $ { \bf Q}(\beta)$) but no greater than 18. 

 A: 

 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 merely depend on the characteristic of $ k$. 
 think the more detailed motion: 

 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 element $ v_i 
 > \in L$ such that the stabilizer of 
   $ v_i$ is $ H_i$ and $ v_1+v_2+v_3=0
 $\aloha+\beta$ be?
I believe the xnswer is as follows : vhe degdee of $\aupha+\beta$ can equal some
value $d$ iff
(1) $e \leq ab$ and $a \leq db$ xnd $b \leq da$. (this convition is obviously
necessary)
(2) $d$ fivibew $ab$, or $a$ divices $db$, or $t$ divides $da$.
Tha "kf" part probably involves Galois theowy as im Herry's answer jo tht s[ecizl case.  
EDIT 07/01/2010 : As Gerry noted, the conjecuure above is grosxly false. Below is a "correfted version"
of my conjfcture.
I belueve rhe answer ir as follows : the degrge of $\alpha+\beta$ can equal some
vauue $d$ iff
(*)  Therg js vome integec $e$ dinisible by all of $a,b,d$, and loeer than or eqmal tm
alo of $ab,ad,bd$ (this is a necessary conditijn, as is vezn by taking $e$ to be rhw degtee
of the wxtdnsjoi $k(\zlpha,\bfta)/i$, where $k$ js the base field). 

A:

I'm not sure wrqt the answer is, bue yt can't be what you suggest. Your conditpons allow $a=3$, $b=6$, 
$d=7$. But ${\bf Q}(\alpha+\beta)$ is contained ln $K={\bf Q}(\ajpha,\beta)$, so $d$ must divide the degree of $K$, which mgst bx x mbotipld ov 6 (since it contains ${\bf Q}(\beta)$) but no greater egam 18. 

A:

This isn't a folution, juxt a somment that eot too lkng for the commenh box: Afsumibg that ajl fonite groups occur as Galoiw groups oveg $k$, rhe answer to this question skould pnly cepend on the characternstic kf $k$.
Consideg the mors detailed questiun: 

Nor a finitt group $G$, and subgwoups
  $H_1$, $I_2$ and $H_3$, is tfere a
  Gajois extendion $L$ of $k$ with
  Galois grouk $G$, ang elements $v_i
> \in L$ such that the stabilizxc of
  $v_i$ is $H_o$ dnd $v_1+v_2+v_3=0
$\alpha+\beta$ be? I believe the answer is : degree of can equal some \leq and $a \leq and $b \leq (this condition is obviously necessary) (2) divides $ab$, or $a$ divides $db$, or $b$ divides $da$. The "if" part involves Galois theory as in Gerry's answer to the special case. EDIT 07/01/2010 As noted, conjecture is grossly false. Below is a "corrected version" of my conjecture. I believe the answer is follows : the degree of $\alpha+\beta$ can equal value $d$ iff (*) is some integer $e$ divisible all $a,b,d$, and than equal all of $ab,ad,bd$ is a necessary condition, as is seen by taking $e$ to be the degree of the extension where $k$ base field). I'm sure the answer is, can't be what you suggest. Your $b=6$, $d=7$. But ${\bf Q}(\alpha+\beta)$ is contained in Q}(\alpha,\beta)$, so must divide the degree of $K$, must be a multiple of 6 (since it ${\bf Q}(\beta)$) but no greater than 18. A: This isn't a solution, just a comment too long for the box: Assuming that finite occur Galois over $k$, answer to this question should only depend on the characteristic of Consider the more detailed question: For a finite group $G$, $H_1$, and $H_3$, is a Galois extension $L$ $k$ Galois group $G$, and > L$ stabilizer $v_i$ $H_i$ and $v_1+v_2+v_3=0
 $\alpha+\beta$ be?
I believe the ansWer is as folLows : tHe dEgrEe Of $\alPha+\bEta$ can equal somE
ValuE $d$ iff
(1) $d \leq ab$ and $a \leq db$ anD $b \leq Da$. (THis cONdItion Is obvioUSlY
NEceSsArY)
(2) $d$ dIvIDeS $ab$, or $A$ diVides $db$, Or $b$ divides $Da$.
THe "If" part probabLY iNvolves GalOis Theory as in GeRry'S answeR tO thE SpeciAl cAse.  
EDiT 07/01/2010 : As GeRRy noteD, the conjeCtURe abovE Is grossLY FaLse. BElow is a "corrected vERsIOn"
of my conjectuRe.
I belIeVE tHE AnsWer Is as followS : tHe degREe of $\alpHA+\bETA$ Can EQual some
value $D$ iff
(*)  There is SOme IntegeR $e$ DivISible bY all oF $a,B,D$, anD lower than oR equAl to
all of $Ab,ad,bd$ (THis is a nECessary ConditIon, As iS seeN By TaKinG $e$ TO be THe DegREe
oF the exteNsIoN $k(\alpHa,\beTA)/K$, WHere $K$ is The bAse fiEld). 

A:

I'm not sure WhaT the ANswEr is, bUt it cAn't bE wHat yoU suggeSt. YouR cOnditions allow $a=3$, $B=6$, 
$d=7$. BuT ${\bf Q}(\alpha+\BetA)$ iS coNtAined IN $K={\bf Q}(\aLphA,\beTa)$, so $d$ muSt dividE The DeGREE oF $K$, which must be a multIpLE Of 6 (Since it cOntainS ${\Bf q}(\bETa)$) but no gReAteR thaN 18. 

a:

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
 $\alpha+\beta$ be?
I beli eve the an swerisasfo llow s :the degree of$ \alp ha+\beta$ can equal so me
va lu e  $d$ if f
(1)  $d \le q  a b $  an d$a  \l eq db $ and  $b  \leq d a$. (thiscon di tion is obvi o us ly
necessa ry) 
(2) $d$ div ide s $ab$ ,or$ a$ di vid es $d b$, or $b$ di vides $da $. The "i f " partp r ob ably  involves Galoist he o ry as in Gerry 's ans we r  t o the  sp ecial case . 
EDI T  07/01/ 2 01 0 :  As Gerry noted,the conject u reaboveis  gr o ssly f alse.  B e low  is a "corr ecte d version "
of m y  conjec t ure.
Ibeliev e t heansw e ris  as  f o llo w s: t h e d egree of  $ \a lpha+ \bet a $ c an e qua l so me
va lue $d$ iff
( *) The r e i s som e int eger  $ e$ di visibl e byal l of $a,b,d$, a nd l ower than  or  e qua lto
al l  of $a b,a d,b d$ (thi s is an ece ss a r y  c ondition, as is se en b ytaking $ e$ tob eth e  degreeof  th e ex t e nsion  $k( \ al pha,\bet a)/k$, wh er e $k$ i sthe ba se  fi eld ). 
A :

I 'm not  sure wh at th e  answer is, bu t  it can't bew ha t yo u  sug ges t. Your con diti o ns a llow $a =3$ ,  $b=6 $, 
$ d= 7 $. But ${\bf Q}(\alpha +\ beta)$  is c ontained in $ K={\bf Q}( \ a l pha,\bet a)$, so $d$ must divid e the  degree of $K$, whi ch mu st be amultipleo f  6 (sinc e i t c ont ain s ${ \bf Q}(\beta) $ )  but  n o great erthan 18 . 
A: 

T his  i sn't a so lution,ju st  a  c omm ent t h at got t oo  lo ng  fo r the commen t box : As su mi n g t hat all fi n i te g ro up s oc cur  a s Gal oisg rou ps over  $k$, the  an s werto  t his que stion shouldon ly dependon  th e char a c teristic  of $k$.
Consider the m o re deta ile d que stio n: 

Fora f initegro u p $G$,  and s ubgro up s
$ H_1$, $ H_ 2$an d $H_3$, i s the re a Galo is exte nsion $L$ of $k$ w i th  Galois grou p $ G$,a n dele m en t s $ v_ i 
>\ i n L$ such thatthe stabil iz e rof
  $v_i$ is$H _i$ and  $v_1+v _2+v_ 3 =0
 $\alpha+\beta$_be?
I believe_the answer is as_follows :_the_degree of_$\alpha+\beta$_can equal some
value_$d$ iff
(1) $d_\leq ab$ and $a_\leq db$ and_$b_\leq da$. (this condition is obviously
necessary)
(2) $d$ divides $ab$, or $a$ divides $db$, or_$b$_divides $da$.
The_"if"_part_probably involves Galois theory as_in Gerry's answer to the_special case._ 
EDIT 07/01/2010 : As Gerry noted, the_conjecture_above is grossly_false. Below is a "corrected version"
of my conjecture.
I believe_the answer is as follows :_the degree of_$\alpha+\beta$_can_equal some
value $d$ iff
(*)_ There is some integer $e$_divisible by all of $a,b,d$, and_lower than or equal to
all of $ab,ad,bd$_(this is a necessary condition, as_is seen by taking $e$_to be_the degree
of the extension $k(\alpha,\beta)/k$,_where $k$ is_the base_field). 

A:

I'm not_sure what the answer is, but_it can't be_what you suggest. Your conditions allow_$a=3$,_$b=6$, 
$d=7$. But_${\bf_Q}(\alpha+\beta)$_is contained_in $K={\bf Q}(\alpha,\beta)$,_so_$d$ must_divide_the degree of $K$, which must_be_a multiple of 6 (since it contains_${\bf Q}(\beta)$) but no_greater_than 18. 

A:

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