(3, 0, -1) = v_{1} - v_{2 }So, (3, 0 -1) is in the span of {v_{1 }, v_{2}}

If (5, 2, -4) is in the span, then (5, 2, -4) = a(1, -2, 3) + b(-2, -2, 4), for some numbers a and b.

So, a - 2b = 5, -2a - 2b = 2, and 3a + 4b = -4

From the second equation, a + b = -1, so b = -1 - a. Therefore, a - 2(-1 - a) = 5. Thus, a = 1 and b = -2. But then, 3a + 4b = -4 is false. So, (5, 2, -4) can't be in the span.

Similarly, if (8, 2, -5) is in the span, then (8, 2, -5) = r(1, -2, 3) + s(-2, -2, 4). This means that r - 2s = 8,

-2r - 2s = 2, and 3r + 4s = -5. So, from the second equation, s = -1 - r. Therefore, r - 2(-1 - r) = 8. So, r = 2 and s = -3. But then 3r + 4s = -5 does not check. So, (8, 2, -5) is not in the spanning set either.

Therefore, the answer is choice (G).