If $\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$ are non-coplanar vectors and $p, q$

are real numbers, then the equality 

$\left[\begin{array}{ccc}3\overrightarrow{u}& p\overrightarrow{v}\hspace{0.25em}& \hspace{0.25em}\hspace{0.25em}p\overrightarrow{w}\end{array}\right]-\left[\begin{array}{ccc}p\overrightarrow{v}& \overrightarrow{w}& q\overrightarrow{u}\end{array}\right]-\left[\begin{array}{ccc}2\overrightarrow{w}& q\overrightarrow{v}& q\overrightarrow{u}\end{array}\right]=0$ holds  for

We have, 

$\left[\begin{array}{l}3\overrightarrow{u}\hspace{0.25em}p\overrightarrow{v}\hspace{0.25em}p\overrightarrow{w}\end{array}\right]-\left[\begin{array}{l}\overrightarrow{p}\overrightarrow{v}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overrightarrow{w}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overrightarrow{qu}\end{array}\right]-\left[\begin{array}{l}2\overrightarrow{w}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overrightarrow{qv}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}q\overrightarrow{u}\end{array}\right]=0$

$\Rightarrow 3p^2\left[\begin{array}{l}\overrightarrow{u}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overrightarrow{v}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overrightarrow{w}\end{array}\right]-pq\left[\begin{array}{l}\overrightarrow{v}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overrightarrow{w}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overrightarrow{u}\end{array}\right]-2q^2\left[\overrightarrow{w}\overrightarrow{v}\overrightarrow{u}\right]=0$

$\Rightarrow 3p^2\left[\begin{array}{ccc}\overrightarrow{u}& \overrightarrow{v}& \overrightarrow{w}\end{array}\right]-pq\left[\begin{array}{ccc}\overrightarrow{u}& \overrightarrow{v}& \overrightarrow{w}\end{array}\right]+2q^2\left[\begin{array}{ccc}\overrightarrow{u}& \overrightarrow{v}& \overrightarrow{w}\end{array}\right]=0$

$\Rightarrow \left(3p^2-pq+2q^2\right)\left[\begin{array}{ccc}\overrightarrow{u}& \overrightarrow{v}& \overrightarrow{w}\end{array}\right]=0$

$\Rightarrow 3p^2-pq+2q^2=0 [\because \left[\begin{array}{ccc}\overrightarrow{u}& \overrightarrow{v}& \overrightarrow{w}\end{array}\right]\ne 0]$

$\Rightarrow p^2-\frac{1}{3}pq+\frac{2}{3}{q}^2=0$

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

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