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Step 1: Expand the first scalar triple product. Using the property $[k_1 \mathbf{a}, k_2 \mathbf{b}, k_3 \mathbf{c}] = k_1 k_2 k_3 [\mathbf{a}, \mathbf{b}, \mathbf{c}]$, we have: $$[3\mathbf{u}, p\mathbf{v}, p\mathbf{w}] = (3)(p)(p) [\mathbf{u}, \mathbf{v}, \mathbf{w}] = 3p^2 [\mathbf{u}, \mathbf{v}, \mathbf{w}]$$ Step 2: Expand the second scalar triple product. Using the same property: $$[p\mathbf{v}, \mathbf{w}, q\mathbf{u}] = (p)(1)(q) [\mathbf{v}, \mathbf{w}, \mathbf{u}]$$ By the cyclic property of scalar triple products, $[\mathbf{v}, \mathbf{w}, \mathbf{u}] = [\mathbf{u}, \mathbf{v}, \mathbf{w}]$. Thus: $$[p\mathbf{v}, \mathbf{w}, q\mathbf{u}] = pq [\mathbf{u}, \mathbf{v}, \mathbf{w}]$$ Step 3: Expand the third scalar triple product. Using the same property: $$[2\mathbf{w}, q\mathbf{v}, q\mathbf{u}] = (2)(q)(q) [\mathbf{w}, \mathbf{v}, \mathbf{u}]$$ To relate $[\mathbf{w}, \mathbf{v}, \mathbf{u}]$ to $[\mathbf{u}, \mathbf{v}, \mathbf{w}]$, we use the property that swapping two vectors changes the sign of the scalar triple product: $[\mathbf{w}, \mathbf{v}, \mathbf{u}] = -[\mathbf{w}, \mathbf{u}, \mathbf{v}]$. Also, $[\mathbf{w}, \mathbf{u}, \mathbf{v}]$ is a cyclic permutation of $[\mathbf{u}, \mathbf{v}, \mathbf{w}]$, so $[\mathbf{w}, \mathbf{u}, \mathbf{v}] = [\mathbf{u}, \mathbf{v}, \mathbf{w}]$. Therefore, $[\mathbf{w}, \mathbf{v}, \mathbf{u}] = -[\mathbf{u}, \mathbf{v}, \mathbf{w}]$. Substituting this back: $$[2\mathbf{w}, q\mathbf{v}, q\mathbf{u}] = 2q^2 (-[\mathbf{u}, \mathbf{v}, \mathbf{w}]) = -2q^2 [\mathbf{u}, \mathbf{v}, \mathbf{w}]$$ Step 4: Substitute all expansions into the given equation. The given equality is $[3\mathbf{u}, p\mathbf{v}, p\mathbf{w}] - [p\mathbf{v}, \mathbf{w}, q\mathbf{u}] - [2\mathbf{w}, q\mathbf{v}, q\mathbf{u}] = 0$. Substituting the expanded forms: $$3p^2 [\mathbf{u}, \mathbf{v}, \mathbf{w}] - (pq [\mathbf{u}, \mathbf{v}, \mathbf{w}]) - (-2q^2 [\mathbf{u}, \mathbf{v}, \mathbf{w}]) = 0$$ $$3p^2 [\mathbf{u}, \mathbf{v}, \mathbf{w}] - pq [\mathbf{u}, \mathbf{v}, \mathbf{w}] + 2q^2 [\mathbf{u}, \mathbf{v}, \mathbf{w}] = 0$$ Step 5: Factor out $[\mathbf{u}, \mathbf{v}, \mathbf{w}]$. $$[\mathbf{u}, \mathbf{v}, \mathbf{w}] (3p^2 - pq + 2q^2) = 0$$ Step 6: Apply the non-coplanarity condition. Since $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are non-coplanar vectors, their scalar triple product $[\mathbf{u}, \mathbf{v}, \mathbf{w}]$ is non-zero. For the equality to hold, the coefficient of $[\mathbf{u}, \mathbf{v}, \mathbf{w}]$ must be zero: $$3p^2 - pq + 2q^2 = 0$$ Step 7: Determine the values of $p$ and $q$. We analyze the quadratic form $3p^2 - pq + 2q^2 = 0$. Case 1: If $q=0$. The equation becomes $3p^2 - p(0) + 2(0)^2 = 0$, which simplifies to $3p^2 = 0$. This implies $p=0$. So, $(p, q) = (0, 0)$ is a solution. Case 2: If $q \neq 0$. We can divide the entire equation by $q^2$: $$3 \left(\frac{p}{q}\right)^2 - \left(\frac{p}{q}\right) + 2 = 0$$ Let $x = \frac{p}{q}$. The equation becomes a quadratic in $x$: $$3x^2 - x + 2 = 0$$ To find the real roots for $x$, we calculate the discriminant $\Delta = b^2 - 4ac$: $$\Delta = (-1)^2 - 4(3)(2) = 1 - 24 = -23$$ Since the discriminant $\Delta < 0$, there are no real solutions for $x$. This means there are no real values for $\frac{p}{q}$ when $q \neq 0$. Combining both cases, the only real solution for $(p, q)$ that satisfies $3p^2 - pq + 2q^2 = 0$ is $(p, q) = (0, 0)$. Therefore, the equality holds for exactly one value of $(p, q)$.