If vec{u}, vec{v}, vec{w} are non-coplanar ve | vedclass

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(C) Given that $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar,their scalar triple product $[\vec{u}, \vec{v}, \vec{w}] 
eq 0$.Using the property of sca

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Exactly one value of $(p, q)$

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(C) Given that $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar,their scalar triple product $[\vec{u}, \vec{v}, \vec{w}] 
eq 0$.Using the property of scalar triple product $[k\vec{a}, l\vec{b}, m\vec{c}] = klm[\vec{a}, \vec{b}, \vec{c}]$,we expand each term:$1$. $[3\vec{u}, p\vec{v}, p\vec{w}] = 3p^2[\vec{u}, \vec{v}, \vec{w}]$$2$. $[p\vec{v}, \vec{w}, q\vec{u}] = pq[\vec{v}, \vec{w}, \vec{u}] = pq[\vec{u}, \vec{v}, \vec{w}]$$3$. $[2\vec{w}, q\vec{v}, q\vec{u}] = 2q^2[\vec{w}, \vec{v}, \vec{u}] = -2q^2[\vec{u}, \vec{v}, \vec{w}]$Substituting these into the equation:$3p^2[\vec{u}, \vec{v}, \vec{w}] - pq[\vec{u}, \vec{v}, \vec{w}] - (-2q^2[\vec{u}, \vec{v}, \vec{w}]) = 0$$(3p^2 - pq + 2q^2)[\vec{u}, \vec{v}, \vec{w}] = 0$Since $[\vec{u}, \vec{v}, \vec{w}] 
eq 0$,we must have $3p^2 - pq + 2q^2 = 0$.For this quadratic form in $p$ and $q$,the discriminant $D = (-q)^2 - 4(3)(2q^2) = q^2 - 24q^2 = -23q^2$.Since $D < 0$ for all $q 
eq 0$,the only real solution is $p = 0, q = 0$. Thus,it holds for exactly one value $(0, 0)$.

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 which of the following?

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