If area is 12 square units then​∣∣​α53​β62​111​∣∣​=±244α−2β−8=±244α−2β=32,4α−2β+16=02α−β−16=0,2α−β+8=0​ Distance from origin​D=α2+(2α+8)2​=5α2+32α+64​(∴β=2α+8)D2=5α2+32α+64​ ​dαd(D2)​=10α+32=0⇒α=−516​β=5−32​+8=58​D=(−516​)2+(58​)2​=58​5​=5​8​​ Similarly if β=2α−16,D=5​16​ So, least possible length of line segment is 5​8​