Question

x^2+y^2+z^2-100/ xy+yz+zx =-2

Answer 1

as (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy +yz+zx)
SO FROM EQ . X.X +Y.Y+ Z.Z -100 + 2(XY+YZ+ZX) =0
SO IT'S (X+Y+Z)^2 =100
AS X+Y = 3Z

SO 16 z.z = 100
Z = 5/4

Answer 2

Answer:

z = $\frac{5}{2}$

Step-by-step explanation:

$x^{2} + y^{2} + z^{2} - 100 = -2 (xy+yz+zx)\\x^{2} + y^{2} + z^{2} - 100 = -2xy -2yz -2zx \\x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx = 100\\\\\\Using (a + b +c )^{2} = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\\\\ (x + y+z)^2 =100\\x+y+z= \sqrt{100} \\x+y+z=10\\\\If x+y=3z, then\\3z+z=10\\4z=10\\z=10/4=5/2\\z=\frac{5}{2}$

Hope the answer is helpful!!

Thanks