If \frac{x^{2}+y^{2}+z^{2}-64}{x y-y z-z x}=-2 and x+y=3 z, then the value of z is (a) 2 (b) 3 (c) 4 (d) None of these

Given the equation \(\frac{x^{2}+y^{2}+z^{2}-64}{xy-yz-zx}=-2\) and the condition that \(x + y = 3z\), we proceed to find the value of \(z\) as follows:

First, we note that:

\[
x^2 + y^2 + z^2 – 64 = -2(xy – yz – zx)
\]

Additionally, we have the identity that can be used:

\[
[x + y + (-z)]^2 = x^2 + y^2 + z^2 + 2(xy – yz – zx)
\]

Substituting \(x + y = 3z\) into this identity, we get:

\[
(3z – z)^2 = x^2 + y^2 + z^2 + 2(xy – yz – zx)
\]

Which simplifies to:

\[
(2z)^2 = x^2 + y^2 + z^2 + 2(xy – yz – zx)
\]

Using the given equation and substituting \(-2(xy – yz – zx)\) for \(x^2 + y^2 + z^2 – 64\), we equate this to \((2z)^2\), resulting in:

\[
(2z)^2 = 64
\]

This simplifies to:

\[
4z^2 = 64
\]

Therefore, solving for \(z\):

\[
z^2 = 16
\]

Given \(z\) is positive (implied by the context), we find:

\[
z = 4
\]

The value of \(z\) is 4.

The answer is (c) 4.