If the expression x2+2(a+b+c)x+3(bc+ca+ab) is a perfect square, then

Given quadratic expression is x2+2(a+b+c)x+3(bc+ca+ab).This quadratic expression will be a perfect square if thediscriminant of its corresponding equation is zero. Hence,4(a+b+c)2−4×3(bc+ca+ab)=0or   (a+b+c)2−3(bc+ca+ab)=0or   a2+b+c2+2ab+2bc+2ca−3(bc+ca+ab)=0or   a2+b2+c2−abbc−ca=0or   122a2+2b2+2c2−2ab−2bc−2ca=0or   12a2+b2−2ab+b2+c2−2bc+c2+a2−2ca=0or  12(a−b)2+(b−c)2+(c−a)2=0which is possible only when (a−b)2=0,(b−c)2=0and (c−a)2=0, i.e., a=b=c