If (x^2 + y^2 + z^2 - 64)/(xy - yz - zx) = –2 and x + y = 3z, then the value of z is

Question

If \(\frac{x^2 + y^2 + z^2 - 64}{xy - yz-zx}\)= –2 and x + y = 3z, then the value of z is

(a) 2 

(b) 3 

(c) 4 

(d) –2

Answer 1

(c) 4 

  \(\frac{x^2 + y^2 + z^2 - 64}{xy - yz-zx}\)= –2 

⇒ x² + y² + z² = - 2xy + 2yz + 2zx + 64

Now,

(x + y + z)² =  x² + y² + z² + 2xy + 2yz + 2zx

⇒ (3z + z)² = - 2xy + 2yz + 2zx + 64 + 2xy + 2yz + 2zx

⇒ 16z² = 4yz + 4xz +64

⇒ 16z² = 4z(x + y) + 64 ⇒ 16z² = 4z (3z) + 64

⇒ 16z² - 12z² = 64

⇒ 4z² = 64 ⇒ z² = 16 ⇒ z = 4.