If $$\vec{u}, \vec{v}, \vec{w}$$ are non-coplanar vectors and $$p, q$$ are real numbers, then the equality $$[3\vec{u} \;\; p\vec{v} \;\; p\vec{w}] - [p\vec{v} \;\; \vec{w} \;\; q\vec{u}] - [2\vec{w} \;\; q\vec{v} \;\; q\vec{u}] = 0$$ holds for

Let $$[\,\vec{a}\;\vec{b}\;\vec{c}\,]$$ denote the scalar triple product $$\vec{a}\cdot(\vec{b}\times\vec{c})$$. It is

(i) linear in every argument     (ii) unchanged by a cyclic permutation     (iii) changes sign when two vectors are interchanged.

Because $$\vec{u},\vec{v},\vec{w}$$ are non-coplanar, $$[\vec{u}\;\vec{v}\;\vec{w}] \neq 0$$.

Evaluate each bracket separately using linearity:

$$[3\vec{u}\; p\vec{v}\; p\vec{w}] = 3p^2\, [\vec{u}\;\vec{v}\;\vec{w}]$$

For the second bracket, perform a cyclic shift $$\vec{v}\rightarrow\vec{w}\rightarrow\vec{u}$$ (no sign change):

$$[p\vec{v}\; \vec{w}\; q\vec{u}] = pq\, [\vec{v}\;\vec{w}\;\vec{u}] = pq\, [\vec{u}\;\vec{v}\;\vec{w}]$$

For the third bracket, interchange the first and third entries of the cyclic set—one transposition introduces a minus sign:

$$[2\vec{w}\; q\vec{v}\; q\vec{u}] = 2q^{2}\,[\vec{w}\;\vec{v}\;\vec{u}] = 2q^{2}\,(-[\vec{u}\;\vec{v}\;\vec{w}]) = -2q^{2}\,[\vec{u}\;\vec{v}\;\vec{w}]$$

Insert these results in the given equality

$$\bigl[3\vec{u}\; p\vec{v}\; p\vec{w}\bigr]\;-\;\bigl[p\vec{v}\; \vec{w}\; q\vec{u}\bigr]\;-\;\bigl[2\vec{w}\; q\vec{v}\; q\vec{u}\bigr]=0$$

$$\Longrightarrow \; \bigl(3p^{2} - pq + 2q^{2}\bigr)\,[\vec{u}\;\vec{v}\;\vec{w}] = 0$$

Since $$[\vec{u}\;\vec{v}\;\vec{w}] \neq 0$$, we require

$$3p^{2} - pq + 2q^{2} = 0 \quad -(1)$$

Equation (1) is a homogeneous quadratic form $$Ap^{2} + 2Bpq + Cq^{2}$$ with $$A=3,\;2B=-1,\;C=2$$. Check its definiteness:

Determinant $$AC-B^{2}=3\cdot2-\left(-\tfrac12\right)^{2}=6-\tfrac14=\tfrac{23}{4}\gt0,\qquad A=3\gt0.$$

Because the determinant is positive and $$A\gt0$$, the form is positive definite. Hence

$$3p^{2} - pq + 2q^{2} \gt 0 \text{ for every } (p,q)\neq(0,0).$$

Therefore equation (1) is satisfied only when $$p=0$$ and $$q=0$$. There is exactly one ordered pair $$(p,q)$$ that makes the original equality true.

Option A which is: exactly one value of $$(p, q)$$