Correct Answer - B
`vec(OG_(1)).vec(BG_(2))=0`
`rArr (a+vecb+vecc)/(3) . (a+vecc-3vecb)/(3)=0`
Now, `|vecc-veca|^(2)=b^(2), |vecc-vecb|=a^(2)` and `|veca-vecb|=c^(2)`.
`therefore 2veca.vecc=a^(2)+c^(2)-b^(2), 2vecb.vecc=b^(2)+c^(2)=a^(2)`,
`2veca.vecb=a^(2)+b^(2)-c^(2)`
Putting in the above result, we get `2a^(2)+2c^(2)-6b^(2)=0`
`rArr (a^(2)+c^(2))/b^(2)=3`