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

exactly one value of (p, q)

We have,

$[3\vec u\, p\vec v\, p\vec w]-[p\vec v\, \vec w\, q\vec u]-[2\vec w\, q\vec v\, q\vec u]=0$

$⇒3p^2[\vec u\, \vec v\, \vec w]-pq[\vec v\, \vec w\, \vec u]-2q^2[\vec w\, \vec v\, \vec u]=0$

$⇒3p^2[\vec u\, \vec v\, \vec w]-pq[\vec u\, \vec v\, \vec w]+2q^2[\vec u\, \vec v\, \vec w]=0$

$⇒(3p^2-pq+2q^2)[\vec u\, \vec v\, \vec w]=0$

$⇒3p^2-pq+2q^2=0$   $[∵[\vec u\, \vec v\, \vec w]≠0]$

$⇒p^2-\frac{1}{3}pq+\frac{2}{3}q^2=0$

$⇒(p-\frac{q}{6})^2+\frac{23}{36}q^2=0⇒p-\frac{q}{6}=0,q=0 ⇒p=0, q=0$

Hence, there is exactly one value of (p, q).