Multiple Choice Question (MCQ)
x = \sqrt{112yz}, y=\sqrt{122xz},z = \sqrt{132xy}
The value of x+y+z is

×
4

✓
6

×
8

×
12
x = \sqrt{112yz}, y=\sqrt{122xz},z = \sqrt{132xy}
The value of x+y+z is
4
6
8
12
x = \sqrt{112yz}, y=\sqrt{122xz},z = \sqrt{132xy}
Square both sides of the equations and addition of them yields
x^2+y^2+z^2 = 11+12+132(yz+xZ+xy)
Then
(x+y+z)^2 = 36
x+y+z = 6
THerefore,
B is the correct option.