All about Ratios
(1) By similar triangles $${OB\over BC}= {3\over 1}{\rm so}
{OB\over OC}={3\over 4}.$$ Again by similar triangles, $${OA\over
OC}= {2\over 3} {\rm so} {OA\over AC}={2\over 5}.$$
The ratio $${AB\over OC} = {OB - OA\over OC} = {3\over 4} - {2\over
5} = {7\over 20}.$$
(2) By similar triangles $${OB\over BC}= {3\over 1}{\rm so}
{OB\over OC}={3\over 4}.$$ Again by similar triangles, $${OA\over
OX}= {4\over 3}$$ where $X$ is the midpoint of $OC$ so $${OA\over
OC}={4\over 14}.$$
The ratio $${AB\over OC} = {OB - OA\over OC} = {3\over 4} - {2\over
7} = {13\over 28}.$$
(3) This is exactly the same as (1) By similar triangles $${OB\over
BC}= {3\over 1} {\rm so} {OB\over OC}={3\over 4}.$$ Again by
similar triangles, $${OA\over OC}= {2\over 3} {\rm so} {OA\over
AC}={2\over 5}.$$
The ratio $${AB\over OC} = {OB - OA\over OC} = {3\over 4} - {2\over
5} = {7\over 20}.$$
Alternatively:
(1) $AB = {7\over 20}\sqrt{10}$, $OC = \sqrt{10}$.
(2) $AB ={13 \sqrt 13\over 28}$, $OC = \sqrt 13$.
(3) $AB ={7\sqrt 13\over 20}$, $OC = \sqrt 13$.