If 𝑢, 𝑣, 𝑤 are non-coplanar vectors and p, q are real numbers, then the equality [3𝑢, p𝑣, p𝑤] − [p𝑣, 𝑤, q𝑢] − [2𝑤, q𝑣, q𝑢] = 0 holds for

We have
3p²[𝑢, 𝑣, 𝑤] − pq[𝑢, 𝑣, 𝑤] − 2q²[𝑢, 𝑣, 𝑤] = 0
Since 𝑢, 𝑣, 𝑤 are non-coplanar vectors, [𝑢, 𝑣, 𝑤] ≠ 0. Therefore, we can divide by [𝑢, 𝑣, 𝑤]:
3p² − pq − 2q² = 0
This is a quadratic equation in p. We can solve for p in terms of q:
p = (q ± √(q² + 24q²)) / 6 = (q ± 5q) / 6
p = q or p = -2q/3
If q = 0, then p = 0. If q = 1, then p = 1 or p = -2/3. If q = 2, then p = 2 or p = -4/3. For every value of q, we have two values of p.
Thus, the equation holds for infinitely many values of (p, q). However, the options don't include this possibility. Let's analyze the quadratic equation further.
The equation 3p² - pq - 2q² = 0 can be factored as (3p + 2q)(p - q) = 0. This gives two solutions for p in terms of q: p = q and p = -2q/3. For each value of q (except q=0), there are two corresponding values of p. If q=0, then p=0. Therefore, there are infinitely many solutions. The closest option is 'More than two but not all values of (p, q)' although it doesn't fully capture the infinite solution set.