Answer to Question #104862 in Algebra for Harishankar

Question

If 3x=3y=3z and x+y+z=27√29

Then √x²+y²+z²=???

Answer 1

Solution. From 3x=3y=3z get x=y=z. As result

$$x=y=z=\frac{x+y+z}{3}=9\sqrt{29}$$

Find the value x²+y²+z²

$$x^2+y^2+z^2=3\times81\times 29$$

Therefore

$$\sqrt {x^2+y^2+z^2}=9\sqrt{87}$$

Answer.

$$\sqrt {x^2+y^2+z^2}=9\sqrt{87}$$