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

  $$\begin{gathered} \left[3upvpw\right]-\left[pvwqu\right]-\left[2wqvqu\right]=0 \Rightarrow 3p^2\left[uvw\right]-pq\left[uvw\right]+2q^2\left[uvw\right]=0 \\ \Rightarrow (3p^2-pq+2q^2)\left[uvw\right]=0 \Rightarrow 3p^2-pq+2q^2=0[\because u,v,warenoncoplanar] \end{gathered}$$
 As a quadratic in p , roots are real $\Rightarrow q^2-24q^2\ge 0\Rightarrow -23q^2\ge 0\Rightarrow q^2\le 0\Rightarrow q=0$
And thus $p=0$  . Thus  $(p,q)\equiv (0,0)$ is the only possibility.