C-03-Q045mediumsingle_mcqIf x+y+z=12x+y+z=12x+y+z=12 and x2+y2+z2=50x^2+y^2+z^2=50x2+y2+z2=50, (x−y)2+(y−z)2+(z−x)2=?(x-y)^2+(y-z)^2+(z-x)^2 = ?(x−y)2+(y−z)2+(z−x)2=?a666b505050c949494d144144144ব্যাখ্যাThe sum equals 2(x2+y2+z2)−2(xy+yz+zx)2(x^2+y^2+z^2)-2(xy+yz+zx)2(x2+y2+z2)−2(xy+yz+zx). Since xy+yz+zx=(x+y+z)2−(x2+y2+z2)2=144−502=47xy+yz+zx=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{144-50}{2}=47xy+yz+zx=2(x+y+z)2−(x2+y2+z2)=2144−50=47, the value is 2(50)−2(47)=62(50)-2(47)=62(50)−2(47)=6.