The problem is to find all integer triples $(a,b,c)$ with $$a<b<c\quad{\rm and} \qquad{1\over a}+{1\over b}+{1\over c}={1\over2}\ .\tag{1}$$ According to your edit you suspect that there could be an infinity of solutions. But this is not the case.

From $(1)$ one immediately infers that ${1\over6}<{1\over a}<{1\over 2}$, which implies $3\leq a\leq 5$. We now treat these cases separately.

(i) When $a=3$ we have to solve $${1\over b}+{1\over c}={1\over 6}$$ with $b<c$, and this implies $7\leq b\leq11$. It follows that $$(b,c)\in\bigl\{(7,42), (8,24), (9,18), (10,15), (11,{\textstyle{5\over66}})\bigr\}\ .$$ This leads to the admissible triples $$(3,7,42), (3,8,24), (3,9,18), (3,10,15)\ .\tag{2}$$ (ii) When $a=4$ we have to solve $${1\over b}+{1\over c}={1\over 4}$$ with $b<c$, and this implies $5\leq b\leq7$. It follows that $$(b,c)\in\bigl\{(5,20), (6,12), (7,{\textstyle{3\over28}})\bigr\}\ .$$ This leads to the admissible triples $$(4,5,20), (4,6,12)\ .\tag{3}$$ (iii) When $a=5$ we have to solve $${1\over b}+{1\over c}={3\over 10}$$ with $b<c$. This implies ${1\over b}>{3\over20}$, or $b\leq6$, and the condition $a<b$ then enforces $b=6$. This would imply $c={15\over2}$, which is not admissble.

All in all there are the $6$ triples listed in $(2)$ and $(3)$.

Since you have already determined all solutions where two or three of the variables are equal it remains to find all integer triples $(a,b,c)$ with $$a<b<c\quad{\rm and} \qquad{1\over a}+{1\over b}+{1\over c}={1\over2}\ .\tag{1}$$ According to your edit you suspect that there could be an infinity of solutions. But this is not the case.

From $(1)$ one immediately infers that ${1\over6}<{1\over a}<{1\over 2}$, which implies $3\leq a\leq 5$. We now treat these cases separately.

(i) When $a=3$ we have to solve $${1\over b}+{1\over c}={1\over 6}$$ with $b<c$, and this implies $7\leq b\leq11$. It follows that $$(b,c)\in\bigl\{(7,42), (8,24), (9,18), (10,15), (11,{\textstyle{5\over66}})\bigr\}\ .$$ This leads to the admissible triples $$(3,7,42), (3,8,24), (3,9,18), (3,10,15)\ .\tag{2}$$ (ii) When $a=4$ we have to solve $${1\over b}+{1\over c}={1\over 4}$$ with $b<c$, and this implies $5\leq b\leq7$. It follows that $$(b,c)\in\bigl\{(5,20), (6,12), (7,{\textstyle{3\over28}})\bigr\}\ .$$ This leads to the admissible triples $$(4,5,20), (4,6,12)\ .\tag{3}$$ (iii) When $a=5$ we have to solve $${1\over b}+{1\over c}={3\over 10}$$ with $b<c$. This implies ${1\over b}>{3\over20}$, or $b\leq6$, and the condition $a<b$ then enforces $b=6$. This would imply $c={15\over2}$, which is not admissble.

All in all there are the $6$ triples listed in $(2)$ and $(3)$.