Algebra problem

Question

If X²+Y²+Z²+294=14(X+Y-2Z),then what is the value of √(XZ-YZ+XY).

Answer 1

Moving everything to the left and completing the square for the x's, y's and z's:

x²-14x+49 +y²-14x+49 + z²+28z+196 [phone] = 0

(x-7)² + (y-7)² + (z+14)² = 0

The only way to add the three groups and get zero is if:

x-7 = 0

y-7 = 0

z+14=0

Therefore:

x=7

y=7

z=-14

√(xz-yz+xy) = √(7(14)-7(14)+7(7)) = √49 = ±7