The least value of the expression x2+4y2+3z2−2x−12y−6z+14 is

Let f(x,y,z)=x2+4y2+3z2−2x−12y−6z+14x2-2x+1-1+4y2-12y+9-9+3z2-6z+3-3+14=(x−1)2+(2y−3)2+3(z−1)2+1 For least value of f(x,y,z)x−1=0;2y−3=0 and z−1=0∴ x=1;y=3/2;z=1 Hence least value of f(x,y,z) is f(1,3/2,1)=1