Proof:

Given: 2^x = 3^y = 6^-z

To show: 1/x * 1/y * 1/z = 0

We can start by expressing 2^x, 3^y, and 6^-z in terms of a common base. Since 6 = 2 * 3, we can rewrite the given equation as:

2^x = (2 * 3)^y = 6^y
3^y = (2 * 3)^y = 6^y
6^-z = (2 * 3)^-z = 2^-z * 3^-z

Now, we can rewrite the equation as:

2^x = 6^y
3^y = 6^y
6^-z = 2^-z * 3^-z

Case 1: x = y
If x = y, then we can substitute the values in the equation:

2^x = 6^x
3^x = 6^x
6^-z = 2^-z * 3^-z

Taking the logarithm of both sides of the equations, we get:

x * log(2) = x * log(6)
x * log(3) = x * log(6)
-z * log(6) = -z * log(2) - z * log(3)

Since x = y, we can divide the equations:

log(2) = log(6)
log(3) = log(6)
log(6) = log(2) + log(3)

Therefore, log(2) = log(3) = log(6). This implies that 2 = 3 = 6, which is not possible. Hence, x = y is not a valid case.

Case 2: x ≠ y
If x ≠ y, we can rewrite the equation as:

2^x = 3^y = 6^(-x-y)

Taking the logarithm of both sides, we get:

x * log(2) = y * log(3) = (-x-y) * log(6)

Dividing the first equation by x and the second equation by y, we have:

log(2) = (y/x) * log(3) = -(1 + y/x) * log(6)

Simplifying further, we get:

log(2) = -(1 + y/x) * log(6)

Dividing both sides by log(2) and rearranging, we have:

1 = -((1 + y/x) * log(6))/log(2)

Since log(6)/log(2) ≠ 0, we can conclude that (1 + y/x) = 0.

Similarly, by dividing the third equation by -z, we can show that 1/z = 0.

Therefore, 1/x * 1/y * 1/z = (1 + y/x) * (1/z) = 0.

Hence, it is proved that 1/x * 1/y * 1/z = 0.