mysql average of best three - mysql

Question

I have pieced this together from sites online and it works but not completely, what i need it to do is take the top 3 results and average them but it takes ALL results, can anyone point me in the right direction?
SELECT i.NAME,
e.comp,
Round(Avg(c.phase1 + c.phase2 + c.phase3 + c.phase4 + c.phase5
+ c.phase6), 2) AS "Average Score",
( CASE
WHEN compID = '7' THEN Concat(Round(Avg(
( (
c.phase1 + c.phase2 + c.phase3 + c.phase4 + c.phase5
+ c.phase6 ) / 400 ) * 100), 2), ' %')
WHEN compID = '5' THEN Concat(Round(Avg(
( (
c.phase1 + c.phase2 + c.phase3 + c.phase4 + c.phase5
+ c.phase6 ) / 600 ) * 100), 2), ' %')
WHEN compID = '3' THEN
Concat(Round(Avg(( ( c.phase1 + c.phase2 + c.phase3 + c.phase4 + c.phase5
+ c.phase6 ) / 600 ) * 100), 2), ' %')
ELSE 'Unspecified'
END ) AS "Average as Percent"
FROM jos_practicedetail c,
jos_comps e,
jos_practice g,
jos_members i
WHERE e.compsid = g.competition
AND g.practiceid = c.practicepid
AND i.memberid = c.competitorid
AND g.typeID = '2'
AND Year(g.pdate) = '2017'
AND (SELECT Count(*)
FROM jos_practicedetail b
WHERE b.competitorid = c.competitorid
AND b.practicepid = c.practicepid
AND ( b.phase1 + b.phase2 + b.phase3 + b.phase4 + b.phase5
+ b.phase6 ) >= (
c.phase1 + c.phase2 + c.phase3 + c.phase4 + c.phase5
+ c.phase6 )) <= 3
GROUP BY competitorid
HAVING Count(*) > 2
ORDER BY competitorid,
( Avg(c.phase1 + c.phase2 + c.phase3 + c.phase4 + c.phase5
+ c.phase6) ) DESC